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7.Binomial Theorem
medium
${(1 + x)^{2n + 1}}$ के विस्तार में महत्तम गुणांक का मान होगा
A
$\frac{{(2n + 1)\,!}}{{n!(n + 1)!}}$
B
$\frac{{(2n + 2)!}}{{n!(n + 1)!}}$
C
$\frac{{(2n + 1)!}}{{{{[(n + 1)!]}^2}}}$
D
$\frac{{(2n)!}}{{{{(n!)}^2}}}$
Solution
$\frac{{{T_{r + 1}}}}{{{T_r}}} = \frac{{N – r + 1}}{r}.x$
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!}}$.
Standard 11
Mathematics
