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7.Binomial Theorem
easy
${\left( {{x^4} - \frac{1}{{{x^3}}}} \right)^{15}}$ के विस्तार में ${x^{32}}$ का गुणांक होगा
A
$^{15}{C_4}$
B
$^{15}{C_3}$
C
$^{15}{C_2}$
D
$^{15}{C_5}$
Solution
${T_{r + 1}} = {}^{15}{C_r}{({x^4})^{15 – r}}{\left( {\frac{{ – 1}}{{{x^3}}}} \right)^r}$
$\therefore {T_{r + 1}} = {}^{15}{C_r}\frac{{{{(x)}^{60 – 4r}}{{( – 1)}^r}}}{{{{(x)}^{3r}}}}$$ = {}^{15}{C_r}{( – 1)^r}{(x)^{60 – 7r}}$
अब $60 – 7r = 32$ रखने पर
==>$60 – 32 = 7r$
==> $r = \frac{{28}}{7} = 4$
$\therefore $ ${r^{32}}$ का गुणांक$ = {}^{15}{C_4}{( – 1)^4} = {}^{15}{C_4}$.
Standard 11
Mathematics
