The number of terms in the expansion of $(1+5\sqrt{2}x)^9$ is (a) 5 (b) 7 (c) 6 (d) 10

The number of terms in the expansion of (1+5√(2)x)⁹ is (a) 5 (b) 7 (c) 6 (d) 10

Question of 64

(1+5\sqrt{2}x)^9

(a) 5

(b) 7

(d) 10

Jharkhand JacJAC Intermediate Board (1st Year) 2023MCQ· 1mImportance★★★★★

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Concept understanding — Binomial Theorem Expansion

The Problem That Started It All

Imagine you have to expand $(x + y)^2$ . That's easy: $x^2 + 2xy + y^2$ . Now try $(x + y)^3$ : $x^3 + 3x^2y + 3xy^2 + y^3$ . Still manageable.

But what about $(x + y)^{10}$ ? Or $(x + y)^{100}$ ? Multiplying it out term by term would take forever. There has to be a pattern — and there is.

The Binomial Theorem is the shortcut that tells you exactly what each term in the expansion of $(x + y)^n$ looks like, without ever having to multiply.

The Pattern You Already Know

Look at the expansions you already know:

Power	Expansion
$(x + y)^0$	$1$
$(x + y)^1$	$x + y$
$(x + y)^2$	$x^2 + 2xy + y^2$
$(x + y)^3$	$x^3 + 3x^2y + 3xy^2 + y^3$
$(x + y)^4$	$x^4 + 4x^3y + 6x^2y^2 + 4xy^3 + y^4$

Notice three things:

The powers of $x$ decrease from $n$ down to $0$ , while the powers of $y$ increase from $0$ up to $n$ . In every term, the exponents add to $n$ .
The coefficients — 1, 4, 6, 4, 1 for $n=4$ — follow a famous pattern called Pascal's triangle.
The number of terms is always $n+1$ .

Note

Pascal's triangle: each number is the sum of the two numbers directly above it.

       1
      1 1
     1 2 1
    1 3 3 1
   1 4 6 4 1
  1 5 10 10 5 1

The Precise Statement

n

Let's break that down.

The symbol $\sum_{k=0}^{n}$ means "add up terms for $k = 0, 1, 2, \dots, n$ ". For each $k$ , the term is:

$\binom{n}{k}$ — the coefficient (read as "n choose k")
$x^{n-k}$ — $x$ raised to the power $n-k$
$y^k$ — $y$ raised to the power $k$

So for $n=4$ , the terms are:

$k$	$\binom{4}{k}$	$x^{4-k}$	$y^k$	Term
0	$\binom{4}{0}=1$	$x^4$	$y^0$	$1 \cdot x^4$
1	$\binom{4}{1}=4$	$x^3$	$y^1$	$4x^3y$
2	$\binom{4}{2}=6$	$x^2$	$y^2$	$6x^2y^2$
3	$\binom{4}{3}=4$	$x^1$	$y^3$	$4xy^3$
4	$\binom{4}{4}=1$	$x^0$	$y^4$	$1 \cdot y^4$

Add them up: $x^4 + 4x^3y + 6x^2y^2 + 4xy^3 + y^4$ . Exactly what we had.

Where Do Those Coefficients Come From?

The binomial coefficient $\binom{n}{k}$ counts how many ways you can choose $k$ items from a set of $n$ items. In the expansion, it counts how many ways you can pick $k$ copies of $y$ (and therefore $n-k$ copies of $x$ ) when multiplying $(x+y)$ by itself $n$ times.

Tip

To compute $\binom{n}{k}$ quickly: start at $n$ and multiply $k$ decreasing numbers, then divide by $k!$ .

Example: $\binom{7}{3} = \frac{7 \cdot 6 \cdot 5}{3 \cdot 2 \cdot 1} = 35$ .

The General Term

The $k$ -th term (starting with $k=0$ ) in the expansion is:

\text{General term: } T_{k+1} = \binom{n}{k} x^{n-k} y^k

This is the most useful part for exams. If someone asks "find the 5th term in $(x+y)^{10}$ ", you set $k=4$ (because $T_{k+1}$ means $k=4$ gives the 5th term) and write:

T_5 = \binom{10}{4} x^{6} y^{4}

What If It's Not Just $x$ and $y$ ? …

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

…

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

Showing the 12 most recent of 31 on this concept.

CBSE 2026Set ANNUAL1 markMCQ
$(x-y)^9$
(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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CBSE 2026Set ANNUAL1 markMCQ
$\displaystyle\sum_{r=0}^{n} 2^r\,{}^nC_r = ?$
(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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CBSE 2026Set ANNUAL1 markMCQ
$(x+a)^n$
(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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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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CBSE 2026Set 1A1 markMCQ
$x^8 y^{10}$
(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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CBSE 2025Set ANNUAL1 markMCQ
$\left(\dfrac{4x}{5} - \dfrac{5}{2x}\right)^9$
(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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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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CBSE 2025Set ANNUAL1 mark
$(x + 2)^6$
›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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CBSE 2024Set ANNUAL1 markMCQ
$\left(x^2-2+\dfrac{1}{x^2}\right)^n$
(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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CBSE 2024Set ANNUAL1 markMCQ
$x^{12}$
(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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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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CBSE 2023Set ANNUAL1 markMCQ
$(1+5\sqrt{2}x)^9$
(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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Q.The number of terms in the expansion of (1+52x)9(1+5\sqrt{2}x)^9(1+52​x)9 is

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

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