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If the coefficient of second, third and fourth terms in the expansion of (1+x)^{2 n} are in A.P. Show that 2 n^{2}-9 n+7=0.
Binomial Theorem | CBSE | Class 11 | Maths | NCERT Exemplar
If the coefficient of second, third and fourth terms in the expansion of (1+x)^{2 n} are in A.P. Show that 2 n^{2}-9 n+7=0.

Given function is (1+x)^{2 n}

{ }^{2 n} C_{1},{ }^{2 n} C_{2} and { }^{2 n} C_{3} are the Coefficient of 2^{\text {nd }}, 3^{\text {rd }} and 4^{\text {th }} terms are respectively.

{ }^{2 n} C_{1},{ }^{2 n} C_{2} and { }^{2 n} C_{3} are in A.P.

So,

2 \cdot{ }^{2 n} C_{2}={ }^{2 n} C_{1}+{ }^{2 n} C_{3}

2\left[\frac{2 n(2 n-1)(2 n-2) !}{2 \times 1 \times(2 n-2) !}\right]=2 n+\frac{2 n(2 n-1)(2 n-2)(2 n-3) !}{3 !(2 n-3) !}

n(2 n-1)=n+\frac{n(2 n-1)(n-1)}{3}

3(2 n-1)=3+\left(2 n^{2}-3 n+1\right)

6 n-3=2 n^{2}-3 n+4 \Rightarrow 2 n^{2}-9 n+7=0

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