7.Binomial Theorem
hard

જો ${(1 + x - 2{x^2})^6} = 1 + {a_1}x + {a_2}{x^2} + .... + {a_{12}}{x^{12}}$, તો ${a_2} + {a_4} + {a_6} + .... + {a_{12}}$ = . . . .

A

$32$

B

$31$

C

$64$

D

એકપણ નહિ.

Solution

(d) ${(1 + x – 2{x^2})^6} = 1 + {a_1}x + {a_2}{x^2} + …. + {a_{12}}{x^{12}}$.

Putting $x = 1$ and $x = -1$ and adding the results $64 = 2(1+a_2+a_4+…)$

$\therefore \,\,\,{a_2} + {a_4} + {a_6} + …. + {a_{12}} = 31$.

Standard 11
Mathematics

Similar Questions

Start a Free Trial Now

Confusing about what to choose? Our team will schedule a demo shortly.