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7.Binomial Theorem
medium
${(1 + x)^{21}} + {(1 + x)^{22}} + .......... + {(1 + x)^{30}}$ के विस्तार में ${x^5}$ का गुणांक होगा
A
$^{51}{C_5}$
B
$^9{C_5}$
C
$^{31}{C_6}{ - ^{21}}{C_6}$
D
$^{30}{C_5}{ + ^{20}}{C_5}$
Solution
${(1 + x)^{21}} + {(1 + x)^{22}} + …. + {(1 + x)^{30}}$
$ = {(1 + x)^{21}}\left[ {\frac{{{{(1 + x)}^{10}} – 1}}{{(1 + x) – 1}}} \right]$= $\frac{1}{x}[{(1 + x)^{31}} – {(1 + x)^{21}}]$
$\therefore$ दिये गये व्यंजक में $x^5$ का गुणांक
= $\left\{ {\frac{1}{x}[{{(1 + x)}^{31}} – {{(1 + x)}^{21}}]} \right\}$ में $x^5$ का गुणांक
= $[{(1 + x)^{31}} – {(1 + x)^{21}}]$ में $x^6$ का गुणांक
= ${}^{31}{C_6} – {}^{21}{C_6}$.
Standard 11
Mathematics
