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Let ( 1 + x + x ^ 2 ) ^ 5 = a 0...

Let ` ( 1 + x + x ^ 2 ) ^ 5 = a _ 0 + a _ 1 x + a _ 2 x ^ 2 + ……. + a _ (10) x ^ ( 10) ` then value of ` a _ 1 + a_ 4 + a _ 7 + a _ (10) ` is ………

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The correct Answer is:
81
`(1+x+x^(2))^(5) = a_(0) + a_(1)x + a_(2)x^(2) + ……… + a_(10) x^(10)`
Put ` x = 1 , omega , omega^(2)`
`a_(0) + a_(1) + a_(2) + ………..+a_(10) = 3^(5) " "…(i)`
`a_(0)+ a_(1)omega + a_(2)omega_(2) + ………….+ a_(10) omega = 0 " "...(ii)`
`a_(0) + a_(1)omega^(2) + a_(2) omega +.........+a_(10) omega^(2) = 0 " "...(iii)`
Multiply (ii) by `omega^(2)` & (iii) by `omega`
> `a_(0) omega^(2) + a_(1) + a_(2) omega + ............ + a_(10) = 0 " "(iv)`
`a_(10) omega + a_(1) + a_(2)omga^(2) + ............+ a_(10) = 0 " "....(v)`
Add (i) + (iv) + (v)
`3(a_(1) + a_(4) + a_(7) + a_(10)) = 3^(5) " " rArr " "a_(1)+ a_(4) + a_(10) = 3^(4) = 81`
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