The gravitational field in a region is given by `vec(g)=(2hat(i)+3hat(j))` N/kg. The work done in moving a particle of mass 1 kg from (1, 1) to `(2, (1)/(3))` along the line 3y + 2x = 5 is

Answer

A, C, D

Explanation

`P(E) = (.^(2n)C_(n))/(2^(2n)) = ((2n)!)/(n! n! 2^(n) 2^(n))`
`=(1xx2xx3xx...xx(2n))/(n!n!2^(n)2^(n))`
`=(1xx3xx5...xx(2n - 1))/(n! 2^(n))`
Now,
`underset(r = 1)overset(n) prod((2r-1)/(2r))=(1xx3xx5xx...xx(2n-1))/(2xx4xx6xx...xx(2n))`
`=(1xx3xx5xx...xx(2n-1))/((1xx2xx3xx...xxn)2^(n))`
`underset(r = 0)overset(n)sum((.^(n)C_(r))/(2^(n)))^(2) = (1)/(2^(n)2^(n)) underset(r = 0)overset(n)sum (.^(n)C_(r))^(2) = (1)/(2^(n)2^(n)) .^(2n)C_(n)`
Also, `(underset(r = 0)overset(n)sum(.^(n)C_(r))^(2))/((underset(r = 0)overset(2n)sum .^(2n)C_(r))) = (.^(2n)C_(n))/(2^(2n))`