The greatest coefficient in the expansion of | vedclass

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Question

(a) $\frac{{{T_{r + 1}}}}{{{T_r}}} = \frac{{N - r + 1}}{r}.x$

Here, $N = 2n +1$==> $\frac{{{T_{r + 1}}}}{{{T_r}}} = \frac{{2n + 2 - r}}{r}.x$

Vedclass · Accepted Answer

$\frac{{(2n + 1)\,!}}{{n!(n + 1)!}}$

Vedclass · Answer

(a) $\frac{{{T_{r + 1}}}}{{{T_r}}} = \frac{{N - r + 1}}{r}.x$

Here, $N = 2n +1$==> $\frac{{{T_{r + 1}}}}{{{T_r}}} = \frac{{2n + 2 - r}}{r}.x$

$	herefore $ ${T_{r + 1}} \ge {T_r}$

$ \Rightarrow $ $2n + 2 - r \ge r$

$ \Rightarrow $ $2n + 2 \ge 2r$ ==> $r \le n + 1$

$	herefore \,\,\,\,\,r = n$${T_{r + 1}} = {T_{n + 1}} = {\,^{2n + 1}}{C_{n + 1}}$

$ = \frac{{(2n + 1)\,!}}{{(n + 1)!\,n!}}$.

The greatest coefficient in the expansion of ${(1 + x)^{2n + 1}}$ is

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