The value of `sum_(r=0)^(20)(-1)^(r )(""^(50)C_(r))/(r+2)` is equal to

Options

A. `(1)/(50xx51)`
B. `(1)/(52xx50)`
C. `1/(52xx51)`
D. none of these

Answer

C

Explanation

Here,
`T_(r)=(-1)^(r)(.^(50)C_(r))/(r+2)`
`=(-1)^(r)(r+1)(.^(50)C_(r))/((r+1)(r+2))`
`=(-1)^(r)(r+1)(.^(52)C_(r+2))/(51xx52)`
`= (-1)^(r)([(r+2)-1]^(52)C_(r+2))/(51xx52)`
`= (-1)^(r)([52.^(51)C_(r+1)-.^(52)C_(r+2)])/(51xx52)`
`= ([-52.^(51)C_(r+1)(-1)^(r+1)-.^(52)C_(r+2)(-1)^(r+2)])/(51xx52)`
`underset(r=0)overset(50)sum(-1)^(r)(.^(50)C_(r))/(r+2)`
` = -52((1-1)^(51)-.^(51)C_(0))/(51xx52)-((1-1)^(52)-.^(52)C_(0)+.^(52)C_(1))/(51xx52)`
`= 1/51-1/52`
`= 1/(51xx52)`
Alternate solution :
`(1-x)^(n)=underset(r=0)overset(n)sum.^(n)C_(r)(-1)^(r)x^(r)`
or `x(1-x)^(n)=underset(r=0)overset(n)sum(-1)^(r).^(n)C_(r)x^(r+1)`
Intergrating both sides withing the limits 0 to 1, we get
`underset(0)overset(1)intx(1-x)^(n)dx=underset(r=0)overset(n)sum(-1)^(r)(.^(n)C_(r))/(r+2)`
or `underset(r=0)overset(n)sum(-1)^(r)(.^(n)C_(r))/(r+2) = underset(0)overset(1)intx(1-x)^(n)dx`
`= underset(0)overset(1)int(1-x)x^(n)dx`(Replace x by `1-x`)
`= |(x^(n+1))/(n+1)-(x^(n+2))/(n+2)|_(0)^(1)`
`= (1)/(n+1)-(1)/(n+2)`
`= (1)/((n+1)(n+2))`
Now put `n = 50`.