Let `Q = sum_(r=0)^(50)(""^(50)C_(r))^(2), R = sum_(r=0)^(100)(-1)^(r) (""^(100)C_(r))^(2)`
Then find the value of Q + R

Answer

C

Explanation

In P, general term of the series is
`T_(r)=(.^(50-r)C_(r)(2r-1))/(.^(50)C_(r)(50+r))`
`=(.^(50+r)C_(r))/(.^(50)C_(r))(1-(50-r+1)/(50+r))`
`= (.^(50+r)C_(r))/(.^(50)C_(r))-(.^(50+r)C_(r))/(.^(50)C_(r))((50-r+1)/(50+r))`
Now,
`(.^(50+r)C_(r))/(.^(50)C_(r))((50-r+1)/(50+r))`
`= ((50-r+1)(50+r)!r!(50-r)!)/(r!50!(50+r)50!)`
`=((50-r+1)!(50+r-1)!)/(50!50!)`
`= (.^(50+r-1)C_(r-1))/(.^(50)C_(r-1))`
`rArr T(r) = (.^(50+r)C_(r))/(.^(50)C_(r)) - (.^(50+r-1)C_(r-1))/(.^(50)C_(r-1))= V(r) - V(r-1)`
Where `V(r) = (.^(50+r)C_(r))/(.^(50)C_(r))`
Now sum of the given series
`P = underset(r=1)overset(50)sumT(r)=V(50)-V(0)`
`= (.^(100)C_(50))/(.^(50)C_(50))- (.^(50)C_(0))/(.^(50)C_(0)) = .^(100)C_(50) - 1`
Also,
`Q = underset(r=0)overset(50)sum(.^(50)C_(r))^(2) = .^(50)C_(0)^(2) + .^(50)C_(1)^(2)+.^(50)C_(2)^(2)+"...."+.^(50)C_(50)^(2)`
`= .^(100)C_(50)`
`rArr P - Q = -1`
We know that
`C_(0)^(2) - C_(1)^(2) + C_(2)^(2) + "....." + (-1)^(n) C_(n)^(2)`
`={{:(0,"If n is odd"),((-1)^(n).^(n)C_(n//2),"if n is even"):}`
`rArr underset(r=0)overset(100)sum(-1)^(r)(.^(100)C_(r))^(2)=(-1)^(100).^(100)C_(50) = .^(100)C_(50)`
`rArr P - R = - 1`
`Q+R = 2.^(100)C_(50) = 2P + 2`