Question

The coefficient of $x^5$ in the expansion of $(1 + x)^{21} + (1 + x)^{22} + ... + (1 + x)^{30}$ is

Answer 1

682017

Answer 2

The problem asks for the coefficient of $x^5$ in the sum $S = (1 + x)^{21} + (1 + x)^{22} + ... + (1 + x)^{30}$.

1. Recognize as a Geometric Progression: The given sum is a geometric progression with first term $a = (1+x)^{21}$, common ratio $r = (1+x)$, and number of terms $N = 30 - 21 + 1 = 10$.
2. Apply GP Sum Formula: The sum $S = \frac{a(r^N - 1)}{r - 1}$ becomes: $S = \frac{(1+x)^{21}((1+x)^{10} - 1)}{(1+x) - 1} = \frac{(1+x)^{31} - (1+x)^{21}}{x}$.
3. Find Coefficient of $x^5$: We need the coefficient of $x^5$ in $\frac{(1+x)^{31}}{x} - \frac{(1+x)^{21}}{x}$. This is equivalent to finding the coefficient of $x^6$ in $(1+x)^{31}$ minus the coefficient of $x^6$ in $(1+x)^{21}$.
4. Use Binomial Theorem: The coefficient of $x^k$ in $(1+x)^n$ is $\binom{n}{k}$. So, the coefficient of $x^5$ in $S$ is $\binom{31}{6} - \binom{21}{6}$.
5. Calculate Binomial Coefficients: $\binom{31}{6} = \frac{31 \times 30 \times 29 \times 28 \times 27 \times 26}{6 \times 5 \times 4 \times 3 \times 2 \times 1} = 736281$. $\binom{21}{6} = \frac{21 \times 20 \times 19 \times 18 \times 17 \times 16}{6 \times 5 \times 4 \times 3 \times 2 \times 1} = 54264$.
6. Subtract to find the final coefficient: $736281 - 54264 = 682017$.