If 1-2x-x 2 n - r-0 2n a r x r- then a r -

The correct option is C ^ 2n C _ r ^ #160; We have #160;( 1+2x+ x ^ 2 ) #160;^ n =∑ _ r=0 ^ 2n a _ r x ^ r #160;⇒( 1+x ) #160;^ 2n =∑ ...

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Factorization Method Form to Remove Indeterminate Form

If 1+2x+x 2...

Question

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

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

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

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^{2 n} C_{r}

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^{2 n} C_{r + 1}

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Solution

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Similar questions

n \geq r - 1

{^{n} C}_{r - 1} + {^{n + 1} C}_{r - 1} + {^{n + 2} C}_{r - 1} + . . . . + {^{2 n} C}_{r - 1} = {^{2 n + 1} C}_{r^{2} - 182} - {^{n} C}_{r}

n

(2 \leq r \leq n)

{^{n} C}_{r} + 2 {^{n} C}_{r + 1} + {^{n} C}_{r + 2}

^{n} C_{r - 1} +^{n + 1} C_{r - 1} +^{n + 2} C_{r - 1} + . . . . . . . +^{2 n} C_{r - 1}, =^{2 n + 1} C_{r^{2} - 132} -^{n} C_{r}

r

n \in N

C_{r} = (\begin{matrix} n r \end{matrix})

0 \leq r \leq n

\sum_{r = 1}^{n} (r) C_{r}^{2} = n (\begin{matrix} 2 n - 1 n - 1 \end{matrix})

\sum_{r = 1}^{n} C_{r}^{2} = (\begin{matrix} 2 n n \end{matrix})

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

C_{0} + \frac{C_{1}}{2} + . . . . + \frac{C_{n}}{n + 1} = \frac{2^{n + 1} - 1}{n + 1}

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