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Question: If \(\sum_{r = 0}^{2n}a_{r}\) (x – 2)<sup>r</sup> = \(\sum_{r = 0}^{2n}b_{r}\) (x – 3)<sup>r</sup> a...

If ∑r=02nar\sum_{r = 0}^{2n}a_{r} (x – 2)r = ∑r=02nbr\sum_{r = 0}^{2n}b_{r} (x – 3)r and ak = 1 for all k ³ n, then bn is equal to –

A

n+1C2n+1

B

2n+1Cn+1

C

2n+1Cn+2

D

None of these

Answer

2n+1Cn+1

Explanation

Solution

Clearly, bn is the coefficient of (x – 3)n in the expression ∑r=02nbr\sum_{r = 0}^{2n}b_{r} (x – 3)r. Therefore,

bn=Coefficient of (x–3)n in(∑r=02nar(x−2)r)\left( \sum_{r = 0}^{2n}{a_{r}(x - 2)^{r}} \right) …(1)

= Coefficient of (x – 3)n in

{∑r=0n−1ar(x−2)r+∑r=n2nar(x−2)r}\left\{ \sum_{r = 0}^{n - 1}{a_{r}(x - 2)^{r} + \sum_{r = n}^{2n}{a_{r}(x - 2)^{r}}} \right\}

= Coefficient of (x – 3)n in

{∑r=0n−1ar(x−2)r+∑r=n2n(x−2)r}\left\{ \sum_{r = 0}^{n - 1}{a_{r}(x - 2)^{r} + \sum_{r = n}^{2n}{(x - 2)^{r}}} \right\}

(Q ak = 1 for all k ³ n)

= Coefficient of (x – 3)n in ∑r=n2n(x−2)r\sum_{r = n}^{2n}{(x - 2)^{r}}

= Coefficient of (x – 3)n in

[(x−2)n{(x−2)n+1−1(x−2)−1}]\left\lbrack (x - 2)^{n}\left\{ \frac{(x - 2)^{n + 1} - 1}{(x - 2) - 1} \right\} \right\rbrack

= Coefficient of (x – 3)n in

{(x−2)2n+1−(x−2)nx−3}\left\{ \frac{(x - 2)^{2n + 1} - (x - 2)^{n}}{x - 3} \right\}

= Coefficient of (x – 3)n+1 in {(x – 2)2n+1 – (x – 2)n}

= Coefficient of (x – 3)n+1 in (x – 2)2n+1

= Coefficient of (x – 3)n+1 in [(x – 3) + 1]2n + 1

= Coefficient of (x – 3)n+1 in

{∑r=02n+12n+1Cr(x−3)r}\left\{ \sum_{r = 0}^{2n + 1}{2n + 1C_{r}(x - 3)^{r}} \right\}

= 2n+1Cn+1.

Hence (2) is correct answer.