STD 11 - 7. binomial theoram — JEE कक्षा 11 साइन्स — Question

यहाँ, $N = 2n +1$ ==> $\frac{{{T_{r + 1}}}}{{{T_r}}} = \frac{{2n + 2 - r}}{r}.x$

$\therefore $   ${T_{r + 1}} \ge {T_r}$

$ \Rightarrow $ $2n + 2 - r \ge r$ $ \Rightarrow $ $2n + 2 \ge 2r$ ==> $r \le n + 1$

$\therefore \,\,\,\,\,r = n$

${T_{r + 1}} = {T_{n + 1}} = {\,^{2n + 1}}{C_{n + 1}}$$ = \frac{{(2n + 1)\,!}}{{(n + 1)!\,n!}}$.