First slide

Binomial theorem for positive integral Index

Question

If $a_{i} (i = 0, 1, 2, \dots, 16)$ be real constants such that for every real value of $x, {(1 + x + x^{2})}^{8} = a_{0} + a_{1} x +$ $a_{2} x^{2} + \dots + a_{16} x^{16},$ then $a_{5}$ is equal to:

Moderate

A

502
B

504
C

506
D

508

Solution

${(1 + x + x^{2})}^{8} = \sum_{\binom{p, q, r \geq 0}{P + q + r = 8}} \frac{8!}{p! q! r!} 1^{p} x^{q} {(x^{2})}^{r}$
For $a_{5}$ , we require, $q + 2 r = 5$
$\begin{array}{l} \Rightarrow q = 5, r = 0, p = 3 or q = 3, r = 1, p = 4 \\ or q = 1, r = 2, p = 5 \end{array}$
Thus, coefficient of $x^{5}$ is
$\frac{8!}{5! 3!} + \frac{8!}{3! 1! 4!} + \frac{8!}{1! 2! 5!} = 504$ .

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