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7.Binomial Theorem
hard
${(1 + 3x + 2{x^2})^6}$ ના વિસ્તરણમાં ${x^{11}}$ નો સહગુણક મેળવો.
A
$144$
B
$288$
C
$216$
D
$576$
Solution
(d) ${(1 + 3x + 2{x^2})^6}$ = ${[1 + x(3 + 2x)]^6}$
= $1 + {\,^6}{C_1}x(3 + 2x){ + ^6}{C_2}{x^2}{(3 + 2x)^2}$${ + ^6}{C_3}{x^3}{(3 + 2x)^3}{ + ^6}{C_4}{x^4}{(3 + 2x)^4}$ ${ + ^6}{C_5}{x^5}{(3 + 2x)^5}{ + ^6}{C_6}{x^6}{(3 + 2x)^6}$
Only ${x^{11}}$ gets from $^6{C_6}{x^6}{(3 + 2x)^6}$
$\therefore $ $^6{C_6}{x^6}{(3 + 2x)^6} = \,{x^6}{(3 + 2x)^6}$
$\therefore $ Coefficient of ${x^{11}}$ = $^6{C_5}{3.2^5} = 576$.
Standard 11
Mathematics
