First slide

Binomial theorem for positive integral Index

Question

If the coefficient of second, third and fourth terms in the expansion of (1 + x)²ⁿ are in AP, then 2n² - 9n is equal to

Moderate

A

-7
B

7
C

-6
D

6

Solution

The general term of $(1 + x)^{2 n} is T_{r + 1} =^{2 n} C_{r} x^{r}$
$T_{2} =^{2 n} C_{1} x^{2}, T_{3} =^{2 n} C_{2} x^{3}, T_{4} =^{2 n} C_{3} x^{4}$
Since , coefficients are in Ap.
$\Rightarrow^{2 n} C_{1},^{2 n} C_{2},^{2 n} C_{3}$ are in AP
$\Rightarrow 2 \times^{2 n} C_{2} =^{2 n} C_{1} +^{2 n} C_{3}$
$\Rightarrow 2 = \frac{^{2 n} C_{1}}{^{2 n} C_{2}} + \frac{^{2 n} C_{3}}{^{2 n} C_{2}}$
$\Rightarrow 2 = \frac{2}{(2 n - 2 + 1)} + \frac{2 n - 3 + 1}{3}$
$\Rightarrow 2 = \frac{2}{2 n - 1} + \frac{2 n - 2}{3}$
$\Rightarrow 2 n^{2} - 9 n + 7 = 0$
$∴ 2 n^{2} - 9 n = - 7$

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