IIT JEE , AIEEE Forum - Mathematics , Algebra

tanmoyoutthere

sum this series --------

S = sigma _{(r=1 to n)}{ (-1)^{r-1 n}C_r /r }

ayshwarya

hey frnd wats d ans plz tell me den i wud post d ans

tanmoyoutthere

the answer was not given ; this was a mcq and none of the answers were matching ---------i came up with the answer
s= (1+1/2+1/3+1/4+....+1/n) * ln 2

this came closest to one of the 4 options which was s= 1+1/2+1/3+...+1/n

tanmoyoutthere

someone solve this plzz

vineetnegi

edited

RyuAmakusa

(1+x)ⁿ = c₀ + c₁x + c₂x²+.......c_nxⁿ.

=> (1+x)ⁿ - 1 = c₁x + c₂x²+.......c_nxⁿ.

=> ((1+x)ⁿ - 1 )/x = c₁ + c₂x+.......c_nx^n-1.

integetate on both sides you get +c sub x=0 and get the val of c. now sub x= -1

and take a - sign common.

vineetnegi

no dear do it completely

i tried it earlier it failed

may you can do itr

i have simple

sum((-1)^r-1*nCr*1/r = sum(-1^r-1((n-1)Cr-1+(n-1)Cr)1/r

break the term

nCr=(n-1)C(r-1)+(n-1)Cr and proceed you will get ans

1/nsum(-1^r-1 (n-1)C(r-1)n/r +sum(-1^r-1n-1Cr*1/r)

it is based on fact

nCr/(n-1)C(r-1)=n/r

nCr=n*[(n-1)C(r-1)*1/r]

1/nsum((-1)^r-1nCr) +SUM(-1^r-1(n-1)Cr*1/r)

1/n+sum(-1^r-1)(n-2)Cr*1/r) + sum(-1^r-1(n-2)C(r-1)*1/r

do same thing

1/n+1/(n-1)sum(1-^r-1(n-1)Cr)+sum(-1^r-1(n-2)Cr)

1/n+1/n-1+sum.......

do it again till n step

1/n+1/n-1+1/n-2

on reversing and writing

1+1/2+1/3..........1/n................ans

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