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જો ${\left( {x + \frac{1}{{{x^2}}}} \right)^{2n}},$ ના વિસ્તરણમાં ${x^m}$ નો સહગુણક મેળવો.
$\frac{{(2n)!}}{{(m)!\,(2n - m)!}}$
$\frac{{(2n)!\,3!\,3!}}{{(2n - m)!}}$
$\frac{{(2n)!}}{{\left( {\frac{{2n - m}}{3}} \right)\,!\,\left( {\frac{{4n + m}}{3}} \right)\,!}}$
એકપણ નહીં.
Solution
(c) ${T_{r + 1}} = {}^{2n}{C_r}{x^{2n – r}}{\left( {\frac{1}{{{x^2}}}} \right)^r}$
= ${}^{2n}{C_r}{x^{2n – 3r}},$
This contains $x^m$, if $2n -3r = m\ i.e$. if $r = \frac{{2n – m}}{3}$
Coefficient of $x^m$ $ = {}^{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)\,\,!}}$
= $\frac{{2n!}}{{\left( {\frac{{4n + m}}{3}} \right)\,\,!\,\,\left( {\frac{{2n – m}}{3}} \right)\,\,!}}$.
