IIT JEE , AIEEE Forum - Mathematics , Algebra

phyana

find the sum of n terms of the series

nC0 + nC3 + nC6 + .............n terms.

plz get me the answer fast with explanation

taruntanuj007

firstly ur question is wrong u cant have nterms since nCr part the lower tersm r in AP

and nth term is 0 + 3(n-1) = 3n-1> n

while writing nCr we have a condition that n >= r and r >= 0 so 0<= r <= n

hence ur question is wrong!!!

truly

(1+x)ⁿ = Co + C₁ x + C₂ x² + C₃X³ + C₄X⁴+.......

let us use cube roots of unity 1,w,w²where 1+w+w² = 0 and w³ = 1
also we have w = -1/2 + i

3 / 2 and w² = -1/2 - i

3 / 2
PUT x = 1
2ⁿ = Co + C₁ + C₂ + C₃ + C₄+.......            eqn1

PUT x=w
(1+w)ⁿ = Co + C₁ w + C₂ w² + C₃w³ + C₄w⁴+.......
(1+w)ⁿ = Co + C₁ w + C₂ w² + C₃ + C₄w+.......         eqn2

PUT x = w²
(1+w²)ⁿ = Co + C₁ w² + C₂ w⁴ + C₃w⁶ + C₄w⁸+.......
(1+w²)ⁿ = Co + C₁ w² + C₂ w + C₃ + C₄w²+.......        eqn3

adding eqn 1 2 and 3

2ⁿ+(1+w)ⁿ+(1+w²)ⁿ = 3(Co+C₃+C₆.........) + (1+w+w²)(C₁+C₂ +C₄+C₅............)
=>2ⁿ+(1+w)ⁿ+(1+w²)ⁿ = 3(Co+C₃+C₆.........)
=>(Co+C₃+C₆.........) = [2ⁿ+(1+w)ⁿ+(1+w²)ⁿ ] / 3

u can now solve further using values of w........I hope its easy.....

phyana

brillant work

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