Q.Find the co-efficient of $x^5$ in the expansion of $(x+3)^8$.

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:

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

   🔒 Start your 14-day free trial to see the full solution →
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:

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

   🔒 Start your 14-day free trial to see the full solution →
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 $r = 0, 1, 2, \dots, n$).

   …

   🔒 Start your 14-day free trial to see the full solution →
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:

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

   …

   🔒 Start your 14-day free trial to see the full solution →
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 $^{18}C_{10}={}^{18}C_8$.

   …

   🔒 Start your 14-day free trial to see the full solution →
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:

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

   …

   🔒 Start your 14-day free trial to see the full solution →
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, $r=3$:

   …

   🔒 Start your 14-day free trial to see the full solution →
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.

   …

   🔒 Start your 14-day free trial to see the full solution →
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: …

   🔒 Start your 14-day free trial to see the full solution →
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 $12$:

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

    …

    🔒 Start your 14-day free trial to see the full solution →
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, $r+1=5 \Rightarrow r=4$:

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

    …

    🔒 Start your 14-day free trial to see the full solution →
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.

    …

    🔒 Start your 14-day free trial to see the full solution →