If the coefficients of 2nd, 3rd and 4th terms in the expansion of (1 + x)²ⁿ are in A.P., show that $2 n^{2} - 9 n + 7 = 0$ .

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Solution

$Given : (1 + x)^{2 n} Thus, we have : T_{2} = T_{1 + 1} = {}^{2 n}C_{1} x^{1} T_{3} = T_{2 + 1} = {}^{2 n}C_{2} x^{2} T_{4} = T_{3 + 1} = {}^{2 n}C_{3} x^{3} We have coefficients of the 2 nd, 3 rd and 4 th terms in AP . ∴ 2 ({}^{2 n}C_{2}) = {}^{2 n}C_{1} + {}^{2 n}C_{3} \Rightarrow 2 = \frac{{}^{2 n}C_{1}}{{}^{2 n}C_{2}} + \frac{{}^{2 n}C_{3}}{{}^{2 n}C_{2}} \Rightarrow 2 = \frac{2}{2 n - 1} + \frac{2 n - 2}{3} \Rightarrow 12 n - 6 = 6 + 4 n^{2} - 4 n - 2 n + 2 \Rightarrow 4 n^{2} - 18 n + 14 = 0 \Rightarrow 2 n^{2} - 9 n + 7 = 0$

Hence proved.

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If the coefficients of 2nd, 3rd and 4th terms in the expansion of $(1 + x)^{2 n}$ are in AP, show that $2 n^{2} - 9 n + 7 = 0$

The coefficients of 2nd, 3rd and 4th terms in the expansion of $(1 + x)^{2 n}$ are in A.P., show that $2 n^{2} - 9 n + 7 = 0.$

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2 n^{2} - 9 n + 7 = 0

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