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If the coefficients of second, third and fourth term in the expansion of  (1+x)^{2 n} are in A.P., then  2 n^2-9 n+7is equal to

 

Option: 1

-1


Option: 2

0


Option: 3

1


Option: 4

3/2


Answers (1)

best_answer

\quad T_2={ }^{2 n} C_1, T_3={ }^{2 n} C_2, T_4={ }^{2 n} C_3  are in A.P. then, 2 \cdot{ }^{2 n} C_2={ }^{2 n} C_1+{ }^{2 n} C_3

2. \frac{2 n(.2 n-1)}{2.1}=\frac{2 n}{1}+\frac{2 n(2 n-1)(2 n-2)}{3.2 .1}

On solving,  2 n^2-9 n+7=0

Posted by

shivangi.bhatnagar

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