IIT JEE , AIEEE Forum - Mathematics , Algebra

sivaram

the find upto n terms of the series

C1 + C5 + C9 ....

or anyother AP of the 'r' in nCr

b_srikalyan009

it is not possible for all A.P's

but for this question here is the idea

(1+x)ⁿ=1+c1x+c2x²+..............

replace x by i i.e

-1

write (i+i) as

2 cis

/4

on write hand side u get R+i(c1-c3+c5-c7....................

here R being the real part

but c1+c3+c5+............=2^{n-1 ....................(1)}

let c1-c3+c5-c7....................=k ^{.......................(2)}

adding 1&2

c1+c5+c9+.............=(k+2^n-1)/2

hence k=2(c1+c5+..........) -- 2^{n-1 ..........................(3)}

R+i(c1-c3+c5.........)=R+i(k)

on right left hand side 2^n/2 * (cos n

/4+isinn

/4)=R+i(k)

hence we get the required sum

rajatsen91

good job srikalyan. This can also be solved by the method of integration which will raise the power of x by 1.

amar.gupta

good work b_srikalyan009

lucky123

correct me if i ma wrong this is a method i cooked up

let S = c1 + c5+ c9 ........

(1+x) ⁿ = c0 + c1x + c2 x^{2 + .......}

let @ = 1^1/4

(1+@)ⁿ = c0 + c1@ +c2 @² + ....

multiply both sides by @³

@³(1+@)ⁿ = S + @ ( c2 + c6+ c10 + ...) + @² (c3+ c7+ c11+ ...) +@³( c0 + c4 + c8 + ..) -------------------(1)

in the above equation

substitute the 4 values for @ ie the four roots of the equation @ = 1^1/4

giving you 4 different equations and all 4

and u will be left with S on RHS ( 1 + @ + @²+@³=0) and some value on LHS

and this method can be generalised for all AP in 'r' for nCr

rate me if u find it useful and correct me if i am wrong

sivaram

so which answer is correct?