First slide

Binomial theorem for positive integral Index

Question

If ${(1 + 2 x + x^{2})}^{n} = \sum_{r = 0}^{2 n} a_{r} x^{r}$ , then a_r =

Very Easy

A

${(^{n} C_{r})}^{2}$
B

$^{n} C_{r} \cdot^{n} C_{r + 1}$
C

$^{2 n} C_{r}$
D

$^{2 n} C_{r + 1}$

Solution

$\begin{array}{l} {(1 + 2 x + x^{2})}^{n} = \sum_{r = 0}^{2 n} a_{r} x^{r} \\ or {[(1 + x)^{2}]}^{n} = \sum_{r = 0}^{2 n} a_{r} x^{r} \\ or (1 + x)^{2 n} = \sum_{r = 0}^{2 n} a_{r} x^{r} \\ or \sum_{r = 0}^{2 n}^{2 n} C_{r} x^{r} = \sum_{r = 0}^{2 n} a_{r} x^{r} \\ or a_{r} =^{2 n} C_{r} \end{array}$

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