If for `n in N ,sum_(k=0)^(2n)(-1)^k(^(2n)C_k)^2=A ,` then find the value of `sum_(k=0)^(2n)(-1)^k(k-2n)(^(2n)C_k)^2dot`

Options

A. `nalpha`
B. `-n alpha `
C. `0
D. none of these

Answer

b

Explanation

Let `beta = sum_(r=0)^(2n) (-1)^(r) (r -2n)(""^(2n)C_(r))^(2)` …(i)
`rArr beta = - sum_(r=0)^(2n) (-1)^(r) (2n -r) (""^(2n)C_(2n-r))^(2)`
Writing the terms in the neverse order, we get
`beta = sum_(r=0)^(2n) (-1)^(r) r(""^(2n)C_(r))^(2) ` ...(ii)
Adding (i) and (ii), we get
`2beta = - sum_(r=0)^(2n) (-1)^(r) (""^(2n)C_(r))^(2)`
`rArr 2 beta = - 2n sum_(r=0)^(2n) (-1)^(r) (""^(2n)C_(r))^(2)`
`rArr 2 beta = - 2nalpha rArr beta = - n alpha ` .