First slide

Binomial theorem for positive integral Index

Question

Let ${(1 + x^{2})}^{2} (1 + x)^{n} = \sum_{r = 1}^{n + 4} a_{r} x^{r}$ . If $a_{1}, a_{2}, a_{3}$ are in A.P., then n is equal to

Moderate

A

6
B

5
C

7
D

2, 3, 4

Solution

$\begin{array}{l} (1 + 2 x^{2} + x^{4}) (1 +^{n} C_{1} x +^{n} C_{2} x^{2} +^{n} C_{3} x^{3} + \dots) \\ = a_{0} + a_{1} x + a_{2} x^{2} + a_{3} x^{3} + \dots \\ \Rightarrow a_{1} =^{n} C_{1}, a_{2} =^{n} C_{2} + 2, a_{3} =^{n} C_{3} + 2 (^{n} C_{1}) \end{array}$
Now, $a_{1}, a_{2}, a_{3}$ are in A.P. implies
$\begin{aligned} 2 (^{n} C_{2} + 2) =^{n} C_{1} +^{n} C_{3} + 2 (^{n} C_{1}) \\ \Rightarrow n (n - 1) + 4 = 3 n + \frac{1}{6} n (n - 1) (n - 2) \\ \Rightarrow 6 (n^{2} - n) + 24 = 18 n + n^{3} - 3 n^{2} + 2 n \\ \Rightarrow n^{3} - 9 n^{2} + 26 n - 24 = 0 \\ \Rightarrow (n - 2) (n - 3) (n - 4) = 0 \Rightarrow n = 2, 3, 4 \end{aligned}$

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