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7.Binomial Theorem
medium
${\left( {x + \frac{1}{{{x^2}}}} \right)^{2n}}$ के विस्तार में ${x^m}$ का गुणांक होगा
A
$\frac{{(2n)!}}{{(m)!\,(2n - m)!}}$
B
$\frac{{(2n)!\,3!\,3!}}{{(2n - m)!}}$
C
$\frac{{(2n)!}}{{\left( {\frac{{2n - m}}{3}} \right)\,!\,\left( {\frac{{4n + m}}{3}} \right)\,!}}$
D
इनमें से कोई नहीं
Solution
${T_{r + 1}} = {}^{2n}{C_r}{x^{2n – r}}{\left( {\frac{1}{{{x^2}}}} \right)^r}$ = ${}^{2n}{C_r}{x^{2n – 3r}},$
2n -3r = m अर्थात् $r = \frac{{2n – m}}{3}$
xm का गुणांक$ = {}^{2n}{C_r},$ $r = \frac{{2n – m}}{3}$
= $\frac{{2n!}}{{(2n – r)!r!}} = \frac{{2n!}}{{\left( {2n – \frac{{2n – m}}{3}} \right)\,\,!\left( {\frac{{2n – m}}{3}} \right)\,\,!}}$
= $\frac{{2n!}}{{\left( {\frac{{4n + m}}{3}} \right)\,\,!\,\,\left( {\frac{{2n – m}}{3}} \right)\,\,!}}$.
Standard 11
Mathematics
