If (1+2x+x2) n=∑r=02narxr, then ar=

We have (1+2x+x2) n=r=0∑2narxr ⇒(1+x) 2n=r=0∑2narxr ⇒r=0∑2n2nCrxr=r=0∑2narxr⇒ar=2nCr

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If { ( 1+2x+x^{ 2 } ) }^{ n }= _{ r=0 }^{ 2n }{ { a }_{ r }{

Question

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

(_{n} C_{r})^{2}

^{n} C_{r} .^{n} C_{r + 1}

^{2 n} C_{r}

^{2 n} C_{r + 1}

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Elizabeth Discussed

(1 + 2 x + x^{2})^{n} = \sum_{r = 0}^{2 n} a_{r} x^{r}

, then

a_{r} =

8 mins ago

Discuss this question LIVE

8 mins ago

We have

(1 + 2 x + x^{2})^{n} = r = 0 \sum 2 n a_{r} x^{r}

\Rightarrow (1 + x)^{2 n} = r = 0 \sum 2 n a_{r} x^{r}

\Rightarrow r = 0 \sum 2 n^{2 n} C_{r} x^{r} = r = 0 \sum 2 n a_{r} x^{r} \Rightarrow a_{r} =^{2 n} C_{r}

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