Q.Show that $9^{n+1} - 8n - 9$ is divisible by 64, whenever $n$ is a positive integer.

Showing the 12 most recent of 15 on this concept.

1. AP EAPCET 2026Set eng-2026-05-14-FN1 markMCQ
   Q.For $n \in N$, $1^2 + 2^2 + 3^2 + \cdots + n^2 >$ (A) $n^3$ (B) $\dfrac{n^3}{2}$ (C) $\dfrac{n^3}{3}$ (D) $3n^3$

   ›Reveal solutionSolution

   The sum of squares $1^2+2^2+\cdots+n^2$ grows like $\frac{n^3}{3}$, so for large $n$ it exceeds $\frac{n^3}{3}$ but is less than $\frac{n^3}{2}$; the only option it is always greater than is $\frac{n^3}{3}$, making (C) the correct choice.

   The key idea is to compare the sum of squares to a simple cubic expression. We know the exact formula:

   $1^2 + 2^2 + \cdots + n^2 = \frac{n(n+1)(2n+1)}{6}.$

   But we don’t need the exact formula to reason about inequalities — we just need to know the leading term. The sum of $k^2$ from $k=1$ to $n$ behaves like $\frac{n^3}{3}$ for large $n$, because the integral $\int_0^n x^2\,dx = n^3/3$ approximates it. So the sum is roughly $n^3/3$, and we can check which of the given options it is always greater than.

   Let’s test each option step by step.

   1. Option (A): $n^3$

      For $n=1$, the sum is $1$, and $n^3=1$, so the sum is not greater (it’s equal). For $n=2$, sum $=1+4=5$, $n^3=8$, so $5 < 8$. So the sum is not always greater than $n^3$. Eliminate (A).

   2. Option (B): $\dfrac{n^3}{2}$

      For $n=1$, sum $=1$, $n^3/2 = 0.5$, so $1 > 0.5$ — okay. For $n=2$, sum $=5$, $n^3/2 = 4$, so $5 > 4$ — okay. For $n=3$, sum $=14$, $n^3/2 = 13.5$, so $14 > 13.5$ — still okay. But check $n=4$: sum $=30$, $n^3/2 = 32$, so $30 < 32$. So it fails for $n=4$. Eliminate (B).

   3. Option (C): $\dfrac{n^3}{3}$

      Test small $n$:

      • $n=1$: sum $=1$, $n^3/3 \approx 0.333$, so $1 > 0.333$.
      • $n=2$: sum $=5$, $n^3/3 \approx 2.667$, so $5 > 2.667$.
      • $n=3$: sum $=14$, $n^3/3 = 9$, so $14 > 9$.
      • $n=4$: sum $=30$, $n^3/3 \approx 21.333$, so $30 > 21.333$. It seems always true. To be sure, use the exact formula:

   $\frac{n(n+1)(2n+1)}{6} > \frac{n^3}{3}$

   Multiply both sides by 6:

   $n(n+1)(2n+1) > 2n^3$

   Expand left: $n(2n^2+3n+1) = 2n^3+3n^2+n$. …

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2. AP EAPCET 2026Set eng-2026-05-14-AN1 markMCQ
   Q.The divisor of $3^{6n} + 56n - 1 \ \forall\ n \in N$ is (A) 729 (B) 625 (C) 676 (D) 784

   ›Reveal solutionSolution

   $3^{6n}+56n-1$ is a classic "$(1+k)^n$ expansion minus a linear correction" divisibility identity; direct evaluation at small $n$ shows it's always a multiple of $784=28^2$.

   Concept and Intuition

   Expressions of the form $a^n + bn - 1$ (here $a=729=3^6$) are divisible by $b^2$ precisely when $b=a-1$, because the binomial expansion $(1+(a-1))^n = 1+n(a-1)+\binom{n}{2}(a-1)^2+\dots$ gives $a^n-1-n(a-1) = \binom{n}{2}(a-1)^2+\dots$, a multiple of $(a-1)^2$. Here $a-1=728$, and $56=728/13$ is NOT quite $a-1$, so it isn't automatically the textbook identity — the safest and most reliable path for an MCQ is to just directly test the divisor candidates against computed values of the expression, which pins down the true common divisor without needing the full expansion argument.

   Step-by-Step Solution

   1. At $n=1$: $3^{6}+56(1)-1 = 729+56-1 = 784$.
   2. At $n=2$: $3^{12}+56(2)-1 = 531441+112-1=531552$. Check $531552/784 = 678$ exactly (since $784\times678=531552$), so $784$ divides this value too.
   3. At $n=3$: $3^{18}+56(3)-1=387420489+168-1=387420656$. Check $387420656/784=494159$ exactly, confirming $784$ divides it again. …

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3. AP EAPCET 2026Set eng-2026-05-18-AN1 markMCQ
   Q.$\forall n \in N$, $\dfrac{n(n+1)^2(n+2)}{12} >$ (A) $\dfrac{(n+2)^4}{12}$ (B) $\dfrac{n^4}{12}$ (C) $\dfrac{n^4}{2}$ (D) $\dfrac{(2n+1)n}{3}$

   ›Reveal solutionSolution

   The inequality $\frac{n(n+1)^2(n+2)}{12} > \frac{n^4}{12}$ holds for all natural numbers $n$, making option (B) the correct choice.

   We are asked: for every natural number $n$, which of the four expressions is always less than the given left-hand side?

   The key is to compare the growth rates and exact values of the expressions, not to solve an equation. Since the inequality must hold for all $n \in \mathbb{N}$, we can test small values and also reason algebraically.

   ---

   Concept & Intuition

   The left-hand side is $\frac{n(n+1)^2(n+2)}{12}$. Notice that for large $n$, the leading term is $\frac{n \cdot n^2 \cdot n}{12} = \frac{n^4}{12}$. So the LHS behaves like $\frac{n^4}{12}$ plus extra positive terms. That suggests the LHS is greater than $\frac{n^4}{12}$ for all $n$. The other options either grow faster (like $\frac{(n+2)^4}{12}$) or are much smaller (like $\frac{(2n+1)n}{3}$), so only one option is plausible.

   ---

   Step-by-step reasoning

   1. Expand the LHS to see its leading term

   $n(n+1)^2(n+2) = n (n^2 + 2n + 1)(n+2)$

   First multiply $(n^2 + 2n + 1)(n+2) = n^3 + 2n^2 + n + 2n^2 + 4n + 2 = n^3 + 4n^2 + 5n + 2$.

   Then multiply by $n$:

   $n^4 + 4n^3 + 5n^2 + 2n$

   So the LHS is $\frac{n^4 + 4n^3 + 5n^2 + 2n}{12}$.

   2. Compare with option (B): $\frac{n^4}{12}$ The difference is:

   $\frac{n^4 + 4n^3 + 5n^2 + 2n}{12} - \frac{n^4}{12} = \frac{4n^3 + 5n^2 + 2n}{12}$

   For any natural number $n \ge 1$, $4n^3 + 5n^2 + 2n > 0$, so the LHS is strictly greater than $\frac{n^4}{12}$.

   Hence option (B) is always true.

   3. Check option (A): $\frac{(n+2)^4}{12}$ Expand $(n+2)^4 = n^4 + 8n^3 + 24n^2 + 32n + 16$. Compare with LHS numerator $n^4 + 4n^3 + 5n^2 + 2n$: The difference (LHS minus option A) is:

   $\frac{(n^4 + 4n^3 + 5n^2 + 2n) - (n^4 + 8n^3 + 24n^2 + 32n + 16)}{12} = \frac{-4n^3 -19n^2 -30n -16}{12}$

   This is negative for all $n \ge 1$, so LHS < option (A). Thus (A) is false.

   4. Check option (C): $\frac{n^4}{2}$ Compare LHS $\frac{n^4 + \dots}{12}$ with $\frac{n^4}{2} = \frac{6n^4}{12}$. The difference is $\frac{n^4 + 4n^3 + 5n^2 + 2n - 6n^4}{12} = \frac{-5n^4 + 4n^3 + 5n^2 + 2n}{12}$. …

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4. AP EAPCET 2025Set eng-2025-05-21-FN1 markMCQ
   Q.For all $n \in N$, if $1^3 + 2^3 + 3^3 + \ldots + n^3 > x$, then a value of $x$ among the following is (A) $\dfrac{n^2}{4}$ (B) $n^2$ (C) $n^4$ (D) $\dfrac{n^2(n+1)^2}{4}$

   ›Reveal solutionSolution

   Using the closed form $\sum k^3 = \dfrac{n^2(n+1)^2}{4}$, only $x=\dfrac{n^2}{4}$ is strictly less than this sum for every natural number $n$.

   Concept and Intuition

   The question asks which expression $x$ can serve as a valid STRICT lower bound of the sum of cubes for every $n$. Since we know the sum exactly, this becomes an algebraic comparison: check, for each option, whether $\text{sum}(n) - x(n) > 0$ for all $n \ge 1$ (not just for large $n$, and not merely $\ge 0$).

   Step-by-Step Solution

   1. Recall the identity: $1^3+2^3+\cdots+n^3 = \left(\dfrac{n(n+1)}{2}\right)^2 = \dfrac{n^2(n+1)^2}{4}$.
   2. Option (D) $\dfrac{n^2(n+1)^2}{4}$: this equals the sum exactly, so "sum $>$ this" is never true (equality, not strict inequality) — rejected.
   3. Option (B) $n^2$: at $n=1$, sum $=1$ and $n^2=1$; sum is NOT $> n^2$ here (equality) — rejected as a universal strict bound.
   4. Option (C) $n^4$: compare $\dfrac{n^2(n+1)^2}{4}$ to $n^4$. Their difference is $\dfrac{n^2(n+1)^2-4n^4}{4} = \dfrac{n^2[(n+1)^2-4n^2]}{4} = \dfrac{n^2(1-n)(3n+1)}{4}$, which is negative for $n\ge2$ (since $1-n<0$ there) — so sum $< n^4$ for $n\ge2$; rejected. …

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5. AP EAPCET 2025Set eng-2025-05-22-FN1 markMCQ
   Q.For all $n \in N$, $\dfrac{3^n-1}{2} \ge$ (A) $n^2\left(2^{n/2}\right)$ (B) $n^2\left(3^{\frac{n-1}{2}}\right)$ (C) $n^3\left(3^{\frac{n-1}{2}}\right)$ (D) $n\left(3^{\frac{n-1}{2}}\right)$

   ›Reveal solutionSolution

   A "which bound is valid" induction-style inequality question — resolved by direct numeric testing across the four candidates, since the exponential-vs-polynomial growth rates make the true bound obvious once you compare a few terms.

   Concept and Intuition

   When four polynomial-times-exponential expressions are offered as lower bounds for $\frac{3^n-1}{2}$, the fastest reliable method is to plug in small $n$ and eliminate any candidate that ever exceeds the left side — because a true lower bound must hold for every $n\in\mathbb{N}$, a single violation kills a candidate.

   Step-by-Step Solution

   1. At $n=1$: LHS $=\dfrac{3-1}{2}=1$. Candidate (A) $=1^2\cdot2^{1/2}\approx1.414 > 1$ — fails already, (A) eliminated.
   2. At $n=2$: LHS $=\dfrac{9-1}{2}=4$. Candidate (B) $=2^2\cdot3^{0.5}\approx6.93>4$ — fails, (B) eliminated. Candidate (C) $=2^3\cdot3^{0.5}\approx13.86>4$ — fails, (C) eliminated.
   3. Candidate (D) at $n=2$: $2\cdot3^{0.5}\approx3.46 \le 4$ — holds.
   4. Check (D) further: $n=3$, LHS $=13$, (D) $=3\cdot3^1=9\le13$ ✓. $n=4$, LHS $=40$, (D) $=4\cdot3^{1.5}\approx20.8\le40$ ✓. $n=5$, LHS$=121$, (D)$=5\cdot3^2=45\le121$ ✓. …

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6. AP EAPCET 2025Set eng-2025-05-24-FN1 markMCQ
   Q.For all $n \in N$, if $n(n^2+3)$ is divisible by $k$, then the maximum value of $k$ is (A) 4 (B) 6 (C) 8 (D) 2

   ›Reveal solutionSolution

   This tests finding the largest $k$ that divides $n(n^2+3)$ for every natural number $n$ — you need the GCD across all $n$, not just one case.

   Concept and Intuition

   The question asks for the greatest common divisor of $n^3+3n$ as $n$ ranges over all naturals. A common trap is to test only one or two values and conclude a bigger number like $4$ or $6$ works — but "divisible by $k$ for all $n$" requires $k$ to divide every instance, so a single counterexample kills a candidate.

   Step-by-Step Solution

   1. Compute $n(n^2+3)$ for the first several naturals:
      • $n=1$: $1(1+3)=4$
      • $n=2$: $2(4+3)=14$
      • $n=3$: $3(9+3)=36$
      • $n=4$: $4(16+3)=76$
      • $n=5$: $5(25+3)=140$
      • $n=6$: $6(36+3)=234$
   2. Any valid $k$ must divide all of these. $\gcd(4,14)=2$. Check the rest: $36/2=18$, $76/2=38$, $140/2=70$, $234/2=117$ — all divide evenly by $2$.
   3. Test whether $4$ works: $14/4=3.5$, not an integer, so $k=4$ fails. …

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7. AP EAPCET 2025Set eng-2025-05-27-FN1 markMCQ
   Q.The remainder obtained when $(2m+1)^{2n}$ $(m,n\in N)$ is divided by $8$ is (A) $1$ (B) $2$ (C) $3$ (D) $4$

   ›Reveal solutionSolution

   The square of any odd integer is always $1$ more than a multiple of $8$ — a standard number-theory fact. Raising that to any further power $n$ keeps the remainder at $1$. Answer: remainder 1.

   Concept and Intuition

   $2m+1$ is the general form of an odd integer. A key fact in elementary number theory: the square of any odd integer, divided by $8$, always leaves remainder $1$. This is because consecutive odd numbers can be paired as $(2m+1)$ and $(2m+1)^2 = 4m^2+4m+1 = 4m(m+1)+1$, and $m(m+1)$ is always a product of two consecutive integers, hence always even — so $4m(m+1)$ is always a multiple of $8$.

   Step-by-Step Solution

   1. Expand $(2m+1)^2 = 4m^2+4m+1 = 4m(m+1)+1$.
   2. $m(m+1)$ is a product of two consecutive integers, so one of them is always even $\Rightarrow$ $m(m+1) = 2k$ for some integer $k$.
   3. So $(2m+1)^2 = 4(2k)+1 = 8k+1$, i.e. $(2m+1)^2 \equiv 1 \pmod 8$.
   4. Now $(2m+1)^{2n} = \left[(2m+1)^2\right]^n \equiv 1^n = 1 \pmod 8$ for every natural number $n$. …

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8. AP EAPCET 2024Set eng-2024-05-18-FN1 markMCQ
   Q.For all positive integers 'n' if $3(5^{2n+1})+2^{3n+1}$ is divisible by k, then the number of prime numbers less than or equal to k is (A) 17 (B) 6 (C) 7 (D) 8

   ›Reveal solutionSolution

   This tests a modular-arithmetic divisibility identity ($3\cdot5^{2n+1}+2^{3n+1}$ is always divisible by 17), followed by counting primes up to that value $k=17$, giving 7 primes.

   Concept and Intuition

   Many 'always divisible by $k$' problems can be cracked cleanly by writing the expression in a form where a common factor emerges directly, rather than relying on induction. Here, expressing $5^{2n+1} = 5\cdot(5^2)^n = 5\cdot 25^n$ and $2^{3n+1} = 2\cdot(2^3)^n = 2\cdot 8^n$, and noting that $25 \equiv 8 \pmod{17}$, lets both terms collapse to multiples of the same power $8^n$, revealing the factor of 17 explicitly.

   Step-by-Step Solution

   1. Rewrite the expression: $3\cdot 5^{2n+1} + 2^{3n+1} = 3\cdot 5 \cdot (5^2)^n + 2\cdot(2^3)^n = 15\cdot 25^n + 2\cdot 8^n$.
   2. Note $25 = 17 + 8$, so modulo 17, $25 \equiv 8$. Hence $25^n \equiv 8^n \pmod{17}$.
   3. Substitute: $15\cdot 25^n + 2\cdot 8^n \equiv 15\cdot 8^n + 2\cdot 8^n = 17\cdot 8^n \equiv 0 \pmod{17}$.
   4. So for every positive integer $n$, the expression is divisible by 17. (Quick check at $n=1$: $3\cdot 5^3+2^4 = 375+16=391 = 17\times 23$ ✓.) …

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9. AP EAPCET 2024Set eng-2024-05-20-AN1 markMCQ
   Q.If $2 \cdot 4^{2n+1} + 3^{3n+1}$ is divisible by $k$ for all $n \in N$, then $k =$ (A) $209$ (B) $11$ (C) $8$ (D) $3$

   ›Reveal solutionSolution

   Rewriting the expression as $8\cdot16^n+3\cdot27^n$ and reducing mod 11 (where $16\equiv27\equiv5$) shows it always equals $11\cdot5^n$, hence always divisible by 11.

   Concept and Intuition

   To find a universal divisor of an expression depending on $n$, it helps to reduce each power modulo the candidate $k$ and see if the combination collapses to a multiple of $k$ for every $n$ — this both proves the divisibility (algebraically, not just by testing values) and identifies exactly which $k$ works.

   Step-by-Step Solution

   1. Rewrite: $2\cdot4^{2n+1}=2\cdot4\cdot4^{2n}=8\cdot(4^2)^n=8\cdot16^n$, and $3^{3n+1}=3\cdot(3^3)^n=3\cdot27^n$.
   2. So the expression is $8\cdot16^n+3\cdot27^n$.
   3. Modulo 11: $16=11+5\equiv5$, and $27=22+5\equiv5$. So $8\cdot16^n+3\cdot27^n\equiv8\cdot5^n+3\cdot5^n=11\cdot5^n\equiv0\pmod{11}$ for all $n$. …

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10. AP EAPCET 2024Set eng-2024-05-22-AN1 markMCQ
    Q.If P is the greatest divisor of $49^{n}+16n-1$ for all $n \in N$, then the number of factors of P is (A) 12 (B) 15 (C) 7 (D) 13

    ›Reveal solutionSolution

    Binomial-expand $49^n=(48+1)^n$ to show $64$ always divides the expression, and the $n=1$ case shows $64$ is exactly the greatest such divisor; $64=2^6$ has $7$ factors.

    Concept and Intuition

    To find the greatest number dividing an expression for every natural number $n$, a standard trick is to rewrite the base as (a multiple of some number) $+1$, so the binomial expansion isolates that number as a common factor of every non-constant term. Here writing $49=48+1$ lets every term except the constant carry a factor of 48 (or a higher power of 48), which combines nicely with the linear $16n$ term.

    Step-by-Step Solution

    1. Write $49^n=(48+1)^n=\sum_{k=0}^n \binom{n}{k}48^k = 1+48n+\binom{n}{2}48^2+\binom{n}{3}48^3+\cdots$
    2. So $49^n+16n-1 = \left(1+48n+\binom{n}{2}48^2+\cdots\right)+16n-1 = 64n+\binom{n}{2}48^2+\binom{n}{3}48^3+\cdots$
    3. Note $48^2=2304=64\times 36$, so $48^2$ (and every higher power of 48) is itself divisible by 64. Hence every term in the sum ($64n$ and all the $48^k$ terms for $k\ge2$) is divisible by 64.
    4. So $64$ divides $49^n+16n-1$ for every $n\in\mathbb{N}$ — 64 is a common divisor. …

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11. AP EAPCET 2021Set eng-2021-08-20-AN1 markMCQ
    Q.Using mathematical induction, the numbers $a_n$'s are defined by $a_0 = 1, a_{n+1} = 3n^2 + n + a_n\ (n \ge 0)$, then $a_n =$ (A) $n^3 + n^2 + 1$ (B) $n^3 - n^2 + 1$ (C) $n^3 - n^2$ (D) $n^3 + n^2$

    ›Reveal solutionSolution

    Generate the first few terms from the recurrence and match them against the candidate closed

    forms; $a_n = n^3-n^2+1$ fits every computed term.

    Concept and Intuition

    When a recurrence defines a sequence and the answer choices are closed forms, the fastest rigorous

    check is to compute several terms directly from the recurrence and test each candidate — the

    correct formula must match every computed value, not just one.

    Step-by-Step Solution

    1. $a_0 = 1$ (given).
    2. $a_1 = a_{0+1} = 3(0)^2 + 0 + a_0 = 0 + 0 + 1 = 1$.
    3. $a_2 = a_{1+1} = 3(1)^2 + 1 + a_1 = 3 + 1 + 1 = 5$.
    4. $a_3 = a_{2+1} = 3(2)^2 + 2 + a_2 = 12 + 2 + 5 = 19$.
    5. Test $a_n = n^3-n^2+1$: at $n=0$: $0-0+1=1$ ✓; at $n=1$: $1-1+1=1$ ✓; at $n=2$: $8-4+1=5$ ✓; at $n=3$: $27-9+1=19$ ✓. All four match. …

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12. AP EAPCET 2021Set eng-2021-08-23-AN1 markMCQ
    Q.If 'n' is a positive integer, then $2\cdot 4^{2n+1}+3^{3n+1}$ is divisible by (A) 2 (B) 9 (C) 11 (D) 27

    ›Reveal solutionSolution

    Checking small cases (and confirming via modular arithmetic) shows $2\cdot4^{2n+1}+3^{3n+1}$ is always divisible by $11$.

    Concept and Intuition

    For "which number always divides this expression" questions, it's efficient to (a) test small values of $n$ to screen out wrong options, and (b) confirm the surviving candidate using modular arithmetic (showing the expression is $\equiv0$ modulo that number for all $n$).

    Step-by-Step Solution

    1. Test $n=1$: $2\cdot4^{2(1)+1}+3^{3(1)+1} = 2\cdot4^3+3^4 = 2(64)+81=128+81=209$. Factor: $209 = 11\times19$. So $209$ is divisible by $11$, but $209$ is odd (not divisible by 2), and $209/9\approx23.2$, $209/27\approx7.7$ (not divisible by 9 or 27).
    2. This already eliminates options (A) 2, (B) 9, (D) 27, leaving (C) 11 as the only surviving candidate.
    3. Confirm with $n=2$: $2\cdot4^5+3^7 = 2(1024)+2187=2048+2187=4235=11\times385$ — again exactly divisible by 11. …

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