If the coefficients of second, third and four | vedclass

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Question

(b) ${T_2} = {}^{2n}{C_1}\,\,x$, ${T_3} = {}^{2n}{C_2}\,\,{x^2}$, ${T_4} = {}^{2n}{C_3}\,\,{x^3}$ 

Coefficient of $T_2, T_3, T_4$ are in $A.P$

Vedclass · Accepted Answer

$0$

Vedclass · Answer

(b) ${T_2} = {}^{2n}{C_1}\,\,x$, ${T_3} = {}^{2n}{C_2}\,\,{x^2}$, ${T_4} = {}^{2n}{C_3}\,\,{x^3}$ 

Coefficient of $T_2, T_3, T_4$ are in $A.P$. 

==> $2.{}^{2n}{C_2} = {}^{2n}{C_1} + {}^{2n}{C_3}$ 

==> $2\frac{{2n!}}{{2\,!\,(2n - 2)\,!}} = \frac{{2n!}}{{(2n - 1)\,!}} + \frac{{2n!}}{{3\,!\,(2n - 3)\,!}}$

==> $\frac{{2\,.\,2n(2n - 1)}}{2} = 2n + \frac{{\,2n(2n - 1)(2n - 2)}}{6}$ 

==> $n(2n - 1) = n + \frac{{(n)(2n - 1)(2n - 2)}}{6}$ 

==> $6(2{n^2} - n) = 6n + 4{n^3} - 6{n^2} + 2n$ 

==> $6n(2n - 1) = 2n(2{n^2} - 3n + 4)$ 

==> $6n - 3 = 2{n^2} - 3n + 4$ 

==> $0 = 2{n^2} - 9n + 7$

==> $2{n^2} - 9n + 7 = 0$.

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

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