Q.Using binomial theorem, find (a + b)^4 - (a - b)^4. Hence evaluate (√3 + √2)^4 - (√3 - √2)^4. OR Find the sum of n terms of the sequence 8, 88, 888, 8888, ..........

Showing the 12 most recent of 31 on this concept.

1. CBSE 2026Set ANNUAL1 markMCQ
   Q.Total number of terms in the expansion of $(x-y)^9$ is

   (a) 9

   (b) 10

   (c) 8

   (d) 11

   ›Reveal solutionSolution

   By the binomial theorem, $(a+b)^n$ (or $(a-b)^n$) always expands into exactly $n+1$ terms, with exponents of $a$ running from $n$ down to $0$.

   The binomial expansion is:

   $(x-y)^n = \sum_{r=0}^{n} {}^nC_r\, x^{n-r}(-y)^r$ …

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2. CBSE 2026Set ANNUAL1 markMCQ
   Q.$\displaystyle\sum_{r=0}^{n} 2^r\,{}^nC_r = ?$

   (a) $4^n$

   (b) $3^n$

   (c) $2^n$

   (d) None of these

   ›Reveal solutionSolution

   Recognise the sum as the binomial expansion $(1+x)^n = \sum {}^nC_r x^r$ with $x=2$.

   The binomial theorem states:

   $(1+x)^n = \sum_{r=0}^{n} {}^nC_r\, x^r$ …

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3. CBSE 2026Set ANNUAL1 markMCQ
   Q.There are 7 terms in the expansion of $(x+a)^n$, then the value of $n$ is —

   (a) $7$

   (b) $6$

   (c) $8$

   (d) $0$

   ›Reveal solutionSolution

   $n = 6$, option (b).

   By the binomial theorem, $(x+a)^n = \sum_{r=0}^{n} \binom{n}{r} x^{n-r} a^r$, which has exactly $n+1$ terms (for $r = 0, 1, 2, \dots, n$).

   …

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4. CBSE 2026Set ANNUAL1 markMCQ
   Q.The number of terms in the expansion of (1 + x)⁴ + (1 - x)⁴ after simplification is:

   (a) 2

   (b) 3

   (c) 4

   (d) 5

   ›Reveal solutionSolution

   Adding (1+x)⁴ and (1-x)⁴ cancels all odd-power terms, leaving only the even-power terms — 3 in total.

   Expand both using the binomial theorem:

   $(1+x)^4 = 1+4x+6x^2+4x^3+x^4$

   $(1-x)^4 = 1-4x+6x^2-4x^3+x^4$

   Adding:

   $(1+x)^4+(1-x)^4 = 2 + 12x^2 + 2x^4 = 2(1+6x^2+x^4)$

   …

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5. CBSE 2026Set 1A1 markMCQ
   Q.The coefficient of $x^8 y^{10}$ in the expansion of $(x + y)^{18}$ is -

   (1) $^{18}C_8$

   (2) $^{18}P_{10}$

   (3) $2^{18}$

   (4) None of these

   ›Reveal solutionSolution

   Coefficient of $x^8y^{10}$ in $(x+y)^{18}$ is $^{18}C_{10}={}^{18}C_8$.

   …

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6. CBSE 2025Set ANNUAL1 markMCQ
   Q.Find the 6th term of the expansion $\left(\dfrac{4x}{5} - \dfrac{5}{2x}\right)^9$.

   (a) $\dfrac{5040}{x}$

   (b) $\dfrac{4050}{x}$

   (c) $\dfrac{-5040}{x}$

   (d) $\dfrac{-4050}{x}$

   ›Reveal solutionSolution

   Using $T_{r+1} = {}^9C_r \left(\dfrac{4x}{5}\right)^{9-r}\left(-\dfrac{5}{2x}\right)^r$ with $r=5$ (for the 6th term) gives $T_6 = \dfrac{-5040}{x}$.

   The general term of $\left(\dfrac{4x}{5} - \dfrac{5}{2x}\right)^9$ is:

   $T_{r+1} = {}^9C_r \left(\frac{4x}{5}\right)^{9-r}\left(-\frac{5}{2x}\right)^r$

   For the 6th term, $r+1=6 \implies r=5$.

   ${}^9C_5 = 126$

   $\left(\frac{4x}{5}\right)^4 = \frac{256x^4}{625}$

   $\left(-\frac{5}{2x}\right)^5 = -\frac{3125}{32x^5}$

   Multiplying:

   $T_6 = 126 \times \frac{256x^4}{625} \times \left(-\frac{3125}{32x^5}\right)$

   …

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7. CBSE 2025Set ANNUAL1 mark
   Q.Find 4th term of the expansion of (x − 2y)^6.

   ›Reveal solutionSolution

   The 4th term corresponds to $r=3$ in the binomial expansion of $(x-2y)^6$.

   For $(x-2y)^6$, the general term is

   $T_{r+1} = {}^6C_r\, x^{6-r}(-2y)^r$

   For the 4th term, $r=3$:

   …

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8. CBSE 2025Set ANNUAL1 mark
   Q.Write the total number of terms in the expansion of $(x + 2)^6$.

   ›Reveal solutionSolution

   The binomial expansion of $(a+b)^n$ has $n+1$ terms; for $n = 6$ this gives $7$ terms.

   By the Binomial Theorem, $(a+b)^n = \displaystyle\sum_{k=0}^{n} {}^nC_k\, a^{n-k}b^k$, which has terms for $k = 0, 1, 2, \ldots, n$ — a total of $n+1$ terms.

   …

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9. CBSE 2024Set ANNUAL1 markMCQ
   Q.The number of terms in the expansion of $\left(x^2-2+\dfrac{1}{x^2}\right)^n$ is

   (a) $2n+1$

   (b) $2n-1$

   (c) $n+1$

   (d) None of these

   ›Reveal solutionSolution

   Rewrite the trinomial base as a perfect square of a binomial first — this turns the problem into a standard binomial expansion with $2n$ as the exponent.

   Notice that:

   $x^2 - 2 + \frac{1}{x^2} = \left(x - \frac{1}{x}\right)^2$

   (since $\left(x-\dfrac1x\right)^2 = x^2 - 2\cdot x \cdot \dfrac1x + \dfrac{1}{x^2} = x^2 - 2 + \dfrac{1}{x^2}$).

   So: …

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10. CBSE 2024Set ANNUAL1 markMCQ
    Q.The coefficient of $x^{12}$ in the expansion of $\left(x^2+\dfrac{1}{x}\right)^{12}$ is

    (a) 495

    (b) 66

    (c) 110

    (d) None of these

    ›Reveal solutionSolution

    Write the general term of the binomial expansion, find which term gives $x^{12}$, then evaluate its coefficient.

    The general term in the expansion of $\left(x^2+\dfrac1x\right)^{12}$ is:

    $T_{k+1} = {}^{12}C_k (x^2)^{12-k}\left(\frac1x\right)^k = {}^{12}C_k \, x^{24-2k}\, x^{-k} = {}^{12}C_k\, x^{24-3k}$

    We want the power of $x$ to be $12$:

    $24 - 3k = 12 \implies 3k = 12 \implies k = 4$

    …

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11. CBSE 2024Set ANNUAL1 markMCQ
    Q.Find 5th term in binomial expansion of (x/3 - 3y)^7:

    (a) 105x^4y^3

    (b) 105x^3y^4

    (c) 105xy^6

    (d) 105x^6y

    ›Reveal solutionSolution

    The 5th term corresponds to $r=4$ in the binomial expansion, giving $105x^3y^4$.

    General term: $T_{r+1} = \binom{7}{r}\left(\dfrac{x}{3}\right)^{7-r}(-3y)^r$.

    For the 5th term, $r+1=5 \Rightarrow r=4$:

    $T_5 = \binom{7}{4}\left(\dfrac{x}{3}\right)^{3}(-3y)^4$

    …

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12. CBSE 2023Set ANNUAL1 markMCQ
    Q.The number of terms in the expansion of $(1+5\sqrt{2}x)^9$ is

    (a) 5

    (b) 7

    (c) 6

    (d) 10

    ›Reveal solutionSolution

    A binomial raised to power $n$ always expands into exactly $n+1$ terms; here $n=9$ gives 10 terms.

    …

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