Use general term Tᵣ₊₁ = ¹⁰Cᵣ xʳ, set r = 5.

Coefficient of x⁵ in (1 + x)¹⁰ — Finding Specific Coefficients

Question

Find the coefficient of $x^5$ in the expansion of $(1 + x)^{10}$ .

Solution — Step by Step

The general term of $(1 + x)^n$ is:

T_{r+1} = \binom{n}{r} x^r

For our expansion, $n = 10$ , so:

T_{r+1} = \binom{10}{r} x^r

We need the power of $x$ to equal 5. From the general term, the power of $x$ is simply $r$ .

So set $r = 5$ .

Substituting $r = 5$ :

T_6 = \binom{10}{5} x^5

Now compute $\binom{10}{5}$ :

\binom{10}{5} = \frac{10!}{5! \cdot 5!} = \frac{10 \times 9 \times 8 \times 7 \times 6}{5 \times 4 \times 3 \times 2 \times 1} = \frac{30240}{120} = 252

The coefficient of $x^5$ in $(1 + x)^{10}$ is 252.

Why This Works

Every term in a binomial expansion comes from choosing how many times we pick the $x$ (versus the 1) across all 10 factors. To get $x^5$ , we pick $x$ from exactly 5 of the 10 brackets — and the number of ways to do that choosing is $\binom{10}{5}$ .

The formula $T_{r+1} = \binom{n}{r} x^r$ packages this logic cleanly. Once you set $r =$ the power you want, the coefficient falls out directly as $\binom{n}{r}$ . No expansion needed, no guessing — just one substitution.

This is why the general term formula is the single most important tool in Binomial Theorem for exams. CBSE asks “find the coefficient of $x^k$ ” almost every year — this approach solves all of them in under 60 seconds.

Alternative Method

We can also use Pascal’s identity directly. The coefficients in $(1+x)^{10}$ are the 10th row of Pascal’s Triangle:

1,\ 10,\ 45,\ 120,\ 210,\ \mathbf{252},\ 210,\ 120,\ 45,\ 10,\ 1

The coefficients of $x^0, x^1, x^2, \ldots, x^{10}$ appear left to right. Counting to the $x^5$ position (6th entry, 0-indexed), we get 252.

For small $n$ (up to 10–12), Pascal’s Triangle is faster than computing $\binom{n}{r}$ by hand. But for JEE Main where $n$ can be 15 or 20, always use the general term formula — don’t rely on the triangle.

Common Mistake

Students confuse the term number with the power of $x$ . They see $T_6$ and write “the coefficient of $x^6$ ” — but $T_6$ means $r = 5$ , so it gives $x^5$ , not $x^6$ . The subscript of $T$ is always one more than the power of $x$ . Whenever you find $r$ , double-check: power of $x$ = $r$ , term number = $r + 1$ .

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

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