Q.Compute $(98)^5$.

Concept and Intuition

To find the highest power of $10$ dividing a huge number without computing it fully, work modulo successive powers of $10$ using the binomial expansions of $(10\mp1)^n$ — only the low-order terms (in $10^0,10^1,10^2,\dots$) survive modulo any given power of $10$.

Step-by-Step Solution

1. $9^{11}=(10-1)^{11}=\sum_{i=0}^{11}\binom{11}{i}10^i(-1)^{11-i}$. Modulo $1000=10^3$, only $i=0,1,2$ survive:
   • $i=0$: $\binom{11}{0}(-1)^{11}=-1$.
   • $i=1$: $\binom{11}{1}\cdot10\cdot(-1)^{10}=110$.
   • $i=2$: $\binom{11}{2}\cdot100\cdot(-1)^9=55\times100\times(-1)=-5500\equiv500\pmod{1000}$.
   • Sum: $-1+110+500=609$.
2. $11^9=(10+1)^9=\sum_{i=0}^{9}\binom{9}{i}10^i$. Modulo $1000$, only $i=0,1,2$ survive:
   • $i=0$: $1$.
   • $i=1$: $\binom{9}{1}\cdot10=90$.
   • $i=2$: $\binom{9}{2}\cdot100=36\times100=3600\equiv600\pmod{1000}$.
   • Sum: $1+90+600=691$.
3. (Cross-check by repeated squaring mod 1000 gives the same $609$ and $691$.)
4. $9^{11}+11^9 \equiv 609+691=1300\equiv300\pmod{1000}$.
5. Since the sum is $\equiv300\pmod{1000}$: it's divisible by $100$ (as $300$ is a multiple of $100$), but not by $1000$ (remainder $300\ne0$).
6. So the greatest $k$ with $10^k$ dividing $9^{11}+11^9$ is $k=2$.

Common Mistakes

• Only checking divisibility by $2$ or $5$ separately instead of computing modulo the full power of $10$.
• Forgetting that a remainder like $300$ mod $1000$ still implies divisibility by $100$ (just not by $1000$) — don't jump to "not divisible by 10 at all."
• Sign errors in the alternating binomial expansion for $(10-1)^{11}$.

ANSWER: B

Concept and Intuition

Since $11=10+1$, powers of $11$ expand into a sum of powers of $10$ weighted by binomial coefficients — a classic trick for writing $11^n$ digit-by-digit. Subtracting $11^2$ just removes the two lowest-order binomial terms of the $12$th-power expansion (adjusted for $11^2$'s own two terms), leaving a clean descending sum.

Step-by-Step Solution

1. $11^{12}=(1+10)^{12}=\sum_{i=0}^{12}\binom{12}{i}10^i$ and $11^2=(1+10)^2=1+20+100$.
2. Subtracting term by term: the $10^0$ terms cancel; the $10^1$ term is $(12-2)\times10=10\times10$; the $10^2$ term is $(66-1)\times100=65\times100$; all higher terms are just $\binom{12}{i}\times10^i$ for $i=3,\dots,12$.
3. So $11^{12}-11^2 = 1\cdot10^{12}+12\cdot10^{11}+66\cdot10^{10}+220\cdot10^{9}+495\cdot10^{8}+792\cdot10^{7}+924\cdot10^{6}+792\cdot10^{5}+495\cdot10^{4}+220\cdot10^{3}+65\cdot10^{2}+10\cdot10$.
4. Divide every coefficient by $200$: $1\times10^{12}/200=5\times10^{9}$; $12\times10^{11}/200=6\times10^{9}$; $66\times10^{10}/200=33\times10^{8}$; $220\times10^{9}/200=110\times10^{7}$; and the last combined term $6500+100=6600$ divides to $33$.
5. Every printed term in the question matches this division by $200$ exactly, so $k=200$.

Common Mistakes

• Trying to factor $11^{12}-11^2$ as $11^2(11^{10}-1)$ and expanding $11^{10}-1$ instead — that expansion doesn't align with the printed powers of $10$ the way the direct $11^{12}$ expansion does.
• Missing that the very first two printed terms ($5\times10^9$ and $6\times10^9$) come from different powers of $10$ ($10^{12}$ and $10^{11}$) both compressing to the same power after division by $200$.

ANSWER: D

Concept and Intuition

Whenever a series' $n$th term is a ratio of a rising product of odd numbers to $c^n\,n!$, suspect the generalized binomial series $(1-x)^{-p}=\sum_{n\ge0}\dfrac{p(p+1)\cdots(p+n-1)}{n!}x^n$. Matching the numerator/denominator pattern pins down $p$ and $x$.

Step-by-Step Solution

1. The $n$th term ($n\ge1$) of the given series is $t_n=\dfrac{3\cdot5\cdot7\cdots(2n+1)}{4^n\,n!}$.
2. For $p=\tfrac32$: $(1-x)^{-3/2}=\displaystyle\sum_{n=0}^{\infty}\dfrac{\tfrac32\cdot\tfrac52\cdots\tfrac{2n+1}2}{n!}x^n=\sum_{n=0}^\infty\dfrac{3\cdot5\cdots(2n+1)}{2^n\,n!}x^n$.
3. Our term carries an extra factor $2^{-n}$ relative to this coefficient, so $t_n$ is exactly this series' coefficient of $x^n$ evaluated at $x=\tfrac12$ (for $n\ge1$).
4. Hence $y=\displaystyle\sum_{n\ge1}t_n=(1-\tfrac12)^{-3/2}-1$, subtracting the $n=0$ term (which equals $1$).
5. $(1-\tfrac12)^{-3/2}=(\tfrac12)^{-3/2}=2^{3/2}=2\sqrt2$, so $y=2\sqrt2-1$, i.e. $y+1=2\sqrt2$.
6. Squaring: $(y+1)^2=8\Rightarrow y^2+2y+1=8\Rightarrow y^2+2y-7=0$.

Common Mistakes

• Forgetting to subtract the $n=0$ term (equal to $1$) when converting a sum starting at $n=1$ into the closed binomial form.
• Misidentifying $p$ — check the first term numerically: $t_1=\tfrac34$ should equal $p\cdot x=\tfrac32\cdot\tfrac12=\tfrac34$. ✓

ANSWER: B

Concept and Intuition

Alternating sums with binomial coefficients and powers are almost always a disguised binomial expansion $(a+b)^m=\sum\binom{m}{k}a^{m-k}b^k$. Spotting the pattern avoids painstaking term-by-term manipulation.

Step-by-Step Solution

1. Write the general term: for $k=0,1,2,\dots,2n$, the term is $\binom{2n}{k}\dfrac{(-10)^k}{81^n}$ — check signs: $k=0$ gives $+1/81^n$, $k=1$ gives $-\binom{2n}{1}\cdot10/81^n$, $k=2$ gives $+\binom{2n}{2}\cdot100/81^n$, matching the given series.
2. So the sum $=\dfrac{1}{81^n}\displaystyle\sum_{k=0}^{2n}\binom{2n}{k}(-10)^k=\dfrac{(1-10)^{2n}}{81^n}=\dfrac{(-9)^{2n}}{81^n}$.
3. Since $2n$ is even, $(-9)^{2n}=9^{2n}=(9^2)^n=81^n$.
4. So the sum $=\dfrac{81^n}{81^n}=1$.

Common Mistakes

• Missing that $(-9)^{2n}$ is positive (even exponent), momentarily worrying about a sign of $(-1)^n$.
• Not recognizing the binomial pattern and trying to sum term-by-term.

ANSWER: C

Concept and Intuition

For $(1+x)^n$ with $x>0$ and non-integer $n$, the general term is $t_{r+1}=\binom{n}{r}x^r=\dfrac{n(n-1)(n-2)\cdots(n-r+1)}{r!}x^r$. Since $x>0$ and $r!>0$, the sign of $t_{r+1}$ is exactly the sign of the product $n(n-1)\cdots(n-r+1)$. This product stays positive as long as every factor $n-j$ (for $j=0,\ldots,r-1$) is positive; it turns negative the first time one factor crosses zero.

Step-by-Step Solution

1. Here $n=27/5=5.4$.
2. The factors in the product for $t_{r+1}$ are $n, n-1, n-2, \ldots, n-(r-1)$.
3. For $r=6$ (i.e. term $t_7$): factors are $5.4, 4.4, 3.4, 2.4, 1.4, 0.4$ — all positive, so $t_7>0$.
4. For $r=7$ (term $t_8$): factors are $5.4,4.4,3.4,2.4,1.4,0.4,-0.6$ — exactly one negative factor ($n-6=-0.6$), so the whole product is negative.
5. Hence $t_8$ is the first negative term, so $k=8$.

Common Mistakes

• Off-by-one errors between the term index $r$ (used in $\binom{n}{r}$) and the term number $t_{r+1}$.
• Forgetting that only ONE factor needs to be negative for the whole product's sign to flip (since all other factors up to that point are positive).

ANSWER: D

Concept and Intuition

In $(x+a)^{15}$, the general term is $T_{r+1}=\binom{15}{r}x^{15-r}a^r$. Consecutive terms compare via the ratio $\dfrac{T_{r+1}}{T_r}=\dfrac{16-r}{r}\cdot\dfrac{a}{x}$ — this ratio decreases as $r$ increases, so terms grow while the ratio exceeds 1 and shrink once it drops below 1. The greatest term sits right where this ratio crosses 1. First we must extract $a/x$ from the given geometric-mean condition on specific terms.

Step-by-Step Solution

1. Write $T_{11}=\binom{15}{10}x^5a^{10}$ (using $r=10$), $T_8=\binom{15}{7}x^8a^7$ ($r=7$), $T_{12}=\binom{15}{11}x^4a^{11}$ ($r=11$).
2. GM condition: $T_{11}^2=T_8\cdot T_{12}$.
3. Substituting: $\binom{15}{10}^2x^{10}a^{20} = \binom{15}{7}\binom{15}{11}\,x^{12}a^{18}$, so $\left(\dfrac{a}{x}\right)^2 = \dfrac{\binom{15}{7}\binom{15}{11}}{\binom{15}{10}^2}$.
4. Values: $\binom{15}{10}=3003$, $\binom{15}{7}=6435$, $\binom{15}{11}=1365$. Factoring: $3003=3\cdot7\cdot11\cdot13$, $6435=3^2\cdot5\cdot11\cdot13$, $1365=3\cdot5\cdot7\cdot13$.
5. Ratio $=\dfrac{3^3\cdot5^2\cdot7\cdot11\cdot13^2}{3^2\cdot7^2\cdot11^2\cdot13^2}=\dfrac{3\cdot25}{7\cdot11}=\dfrac{75}{77}$.
6. So $a/x=\sqrt{75/77}\approx0.9869$.
7. Term ratio: $\dfrac{T_{r+1}}{T_r}=\dfrac{16-r}{r}\cdot\dfrac{a}{x}$. At $r=7$: $\dfrac{9}{7}\times0.9869\approx1.269>1 \Rightarrow T_8>T_7$.
8. At $r=8$: $\dfrac{8}{8}\times0.9869\approx0.987<1 \Rightarrow T_9<T_8$.
9. Since $T_8$ exceeds both its neighbours, $T_8$ (the 8th term) is the greatest.

Common Mistakes

• Off-by-one errors relating "$r$-th term used in GM" to the exponent $r$ in $\binom{15}{r}$ (the $k$-th term is $T_k=\binom{15}{k-1}x^{16-k}a^{k-1}$).
• Assuming the greatest term must fall at a "nice" integer ratio value; here $(a/x)^2=75/77$ is not a perfect square, but the greatest-term test still works fine via the ratio inequality.

ANSWER: B

Concept and Intuition

The ratio of consecutive binomial coefficients $\dfrac{{}^nC_k}{{}^nC_{k-1}}=\dfrac{n-k+1}{k}$ is a standard identity, which lets every term in the sum collapse to something very simple.

Step-by-Step Solution

1. General term: $k\cdot\dfrac{C_k}{C_{k-1}}=k\cdot\dfrac{n-k+1}{k}=n-k+1$.
2. So the sum $\displaystyle\sum_{k=1}^n k\dfrac{C_k}{C_{k-1}}=\sum_{k=1}^n (n-k+1)$.
3. As $k$ runs $1,2,\ldots,n$, the quantity $(n-k+1)$ runs $n,n-1,\ldots,1$ — the same set of values, just in reverse order.
4. So the sum equals $\displaystyle\sum_{m=1}^{n} m=\sum_{k=1}^n k$.

Common Mistakes

• Trying to compute each ratio term explicitly instead of recognizing the ratio identity, leading to unnecessary algebra.
• Confusing $\sum k$ with $\sum k^2$ (option A) by mis-simplifying the general term.

ANSWER: D

Concept and Intuition

When you expand $(x+y)^5$ and $(x-y)^5$ by the binomial theorem, terms with an odd power of $y$ appear with opposite signs in the two expansions (since $(-y)^{\text{odd}}=-y^{\text{odd}}$), so they cancel on addition. Only even powers of $y$ survive, and they double. This is exactly how you extract a rational value out of a sum of two conjugate surds raised to a power.

Step-by-Step Solution

1. Write $x=3,\ y=\sqrt8$. Then

(the $y^1,y^3,y^5$ terms cancel).

2. Compute each piece: $x^5=3^5=243$; $\binom{5}{2}=10,\ x^3=27,\ y^2=8\Rightarrow 10\cdot27\cdot8=2160$; $\binom{5}{4}=5,\ x=3,\ y^4=64\Rightarrow5\cdot3\cdot64=960$.

3. Sum inside the bracket: $243+2160+960=3363$.