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7.Binomial Theorem
easy
${\left( {{x^4} - \frac{1}{{{x^3}}}} \right)^{15}}$ ના વિસ્તરણમાં ${x^{39}}$ નો સહગુણક મેળવો.
A
$-455$
B
$-105$
C
$105$
D
$455$
Solution
(a) ${T_{r + 1}} = \,{}^{15}{C_r}{({x^4})^{15 – r}}{( – 1/{x^3})^r}$= ${( – 1)^r}\,\,{}^{15}{C_r}{(x)^{60 – 7r}}$
For coefficient of ${x^{39}}$, $60 – 7r = 39$
$\Rightarrow r = 3$
$\therefore {T_4} = {}^{15}{C_3}{({x^4})^{12}}{( – 1/{x^3})^3}$= $ – 455\,{x^{39}}$
Hence the required coefficient is $-455$.
Standard 11
Mathematics
