- Home
- Standard 11
- Mathematics
7.Binomial Theorem
hard
If ${(1 + x - 2{x^2})^6} = 1 + {a_1}x + {a_2}{x^2} + .... + {a_{12}}{x^{12}}$, then the expression ${a_2} + {a_4} + {a_6} + .... + {a_{12}}$ has the value
A
$32$
B
$31$
C
$64$
D
None of these
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
