Find coefficient of x⁵ in expansion of (1+x)¹⁰

Question

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

Solution — Step by Step

The general term ( $(r+1)$ th term) in the expansion of $(1 + x)^n$ is:

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

For $(1+x)^{10}$ , we have $n = 10$ .

We need the term containing $x^5$ , so we set $r = 5$ :

T_{5+1} = T_6 = \binom{10}{5} x^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 6th term in the expansion is $252x^5$ .

The coefficient of $x^5$ is 252.

Why This Works

The Binomial Theorem tells us that $(1+x)^n = \sum_{r=0}^{n} \binom{n}{r} x^r$ . Each term $\binom{n}{r} x^r$ gives the coefficient of $x^r$ directly as $\binom{n}{r}$ .

The binomial coefficient $\binom{10}{5} = 252$ counts the number of ways to choose 5 positions (out of 10 brackets) to contribute an $x$ , while the remaining 5 brackets contribute 1. This combinatorial interpretation makes intuitive sense.

Alternative Method — Pascal’s Triangle

The coefficients of $(1+x)^{10}$ are the 11th row of Pascal’s triangle. The 6th entry (starting from 1 at position 0) is 252. For lower powers of $n$ , Pascal’s triangle is faster than computing $\binom{n}{r}$ .

For $\binom{10}{5}$ , note the symmetry: $\binom{10}{5} = \binom{10}{10-5} = \binom{10}{5}$ . For large values, always check if it’s easier to compute from the smaller side. $\binom{10}{5}$ and $\binom{100}{95}$ take the same conceptual effort — always use $\binom{100}{5}$ (the smaller version).

Common Mistake

Students sometimes write $T_r$ instead of $T_{r+1}$ for the general term, leading to an off-by-one error. The standard convention is that the first term is $T_1$ (when $r = 0$ ), the second term is $T_2$ (when $r = 1$ ), and so on. So $x^5$ corresponds to $r = 5$ and $T_6$ — not $T_5$ . Always double-check by expanding the first few terms manually.

Question

Solution — Step by Step

Why This Works

Alternative Method — Pascal’s Triangle

Common Mistake

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