The value of ∑r=050 (−1)r50crr+2 is equal to

Here, Tr=(−1)r 50Crr+2=(−1)r(r+1) 50Cr(r+1)(r+2)=(−1)r(r+1) 52Cr+251×52=(−1)r[(r+2)−1]52Cr+251×52=(−1)r5251Cr+1−52Cr+251×52=−5251Cr+1(−1)r+1−52Cr+2(−1)r+251×52∑r=050 (−1)r 50Crr+2=∑r=050 −5251Cr+1(−1)r+1−52Cr+2(−1)r+251×52=−52(1−1)51−51C051×52−(1−1)52−52C0+52C151×52 =151−152=151×52Alternate solution:(1−x)n=∑r=0n nCr(−1)r xror  x(1−x)n=∑r=0n (−1)r  nCr xr+1Integrating both sides within the limits 0 to 1, we get∫01 x(1−x)ndx=∑r=0n (−1)r nCrr+2or   ∑r=0n (−1)r nCrr+2=∫01 x(1−x)ndx=∫01 (1−x)xndx ( Replace x by 1−x)=xn+1n+1−xn+2n+201=1n+1−1n+2=1(n+1)(n+2)Now put n = 50.