Suppose - r-0 2n a r x-2 r - r-0 2n b r x-3 rfind bn if for each k- n

Putting x 3=y ∑ _ r=0 ^ 2n b _ r y _ r =∑ _ r=0 ^ 2n b _ r (y+1) ^ r ⇒ b _ n = a _ n + _ #160;^ n+1 C _ 1 a _ n+1 + _ #160;^ n+2 C _ ...

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Binomial Coefficients

Suppose ∑ r...

Question

Suppose $\sum_{r = 0}^{2 n} a_{r} {(x - 2)}^{r} = \sum_{r = 0}^{2 n} b_{r} {(x - 3)}^{r}$
find $b_{n}$ if for each $k \geq n$

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\sum_{r = 0}^{2 n} a_{r} {(x - 100)}^{r} = \sum_{r = 0}^{2 n} b_{r} {(x - 101)}^{r}

a_{k} = \frac{2^{k}}{{^{k} C}_{n}}

k \geq n

b_{n}

n \in N

\sum_{k = 0}^{2 n} {(- 1)}^{k} {({^{2 n} C}_{k})}^{2} = A

\sum_{k = 0}^{2 n} {(- 1)}^{k} (k - 2 n) {({^{2 n} C}_{k})}^{2}

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

a_{r} =

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

a_{1} - 2 a_{2} + 3 a_{3} - \dots . - 2 n a_{2 n} = \dots

lim n \to \infty \frac{1}{n} 2 n - 1 \sum r = 0 \frac{n^{2}}{n^{2} + 4 r^{2}}

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