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Prove that (C1)/1-(C2)/2+(C3)/3-(C4)/4+....

Prove that `(C_1)/1-(C_2)/2+(C_3)/3-(C_4)/4+....+((-1)^(n-1))/n C_n=1+1/2+1/3+...+1/n`

Text Solution

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We have,
`(1+x)^(n)=.^(n)C_(0)+.^(n)C_(1)x+.^(n)C_(2)x^(2)+"....."+.^(n)C_(n)x^(n)`
Dividing by x, we get
`((1-x)^(n)-1)/(x) = .^(n)C_(1) + .^(n)C_(2)x+"..... "+.^(n)C_(n)x^(n-1)`
`:. 1+(1+x)+(1+x)^(2) + "...."+(1+x)^(n-1)`
`= .^(n)C_(1) + .^(n)C_(2)x + "..." + .^(n)C_(n)x^(n-1)`.
`:. underset(-1)overset(0)int(.^(n)C_(1) + .^(n)C_(2)x+C_(3)x^(2) + "...."+.^(n)C_(n) x^(n-1))dx`
`= underset(-1)overset(0)int[1+(1+x)+"....."+(1+x)^(n-1)]dx`
`rArr [.^(n)C_(1)x+(.^(n)C_(2)x^(2))/(2)+(.^(n)C_(3)x^(2))/(3)+"...."+(.^(n)C_(n)x^(n))/(n)]_(-1)^(0)`
`= [x+((1+x)^(2))/(2)+((1+x)^(3))/(3)+"....."+((1+x)^(n))/(n)]_(-1)^(0)`
`rArr (.^(n)C_(1))/(1)-(.^(n)C_(2))/(2)+(.^(n)C_(3))/(3)-"...."+((-1)^(n-1))/(n).^(n)C_(n)=1+1/2+1/3+"......"+1/n`
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