If `2^3+4^3+6^3+.....+(2n)^3=kn^2(n+1)^2` then find the value of `k`

If 2^(3)+4^(3)+6^(3)+.. .+(2 n)^(3)=k n^(2)(n+1)^(2) then k=

If 2^(3) + 4^(3) + 6^(3) + … + (2n)^(3) = kn^(2) ( n+1)^(2) then k=

If .^(n)C_(3)=k.n(n-1)(n-2) then k=

Sum of the series 1+(1+2)+(1+2+3)+(1+2+3+4)+... to n terms be 1/m n(n+1)[(2n+1)/k+1] then find the value of k+m

If m = 3 and n = 2 find the value of 2m^3 +3n^2- 5 m^2n

(i) If (n!)/(2.(n-2)!): (n!)/(4!.(n-4)!) = 2:1 , find the valye of n. (ii) If ((2n)!)/(3!(2n-3)!): (n!)/(2!(n-2)!) = 44:3 , then find the value of n.

(i) If (n!)/(2.(n-2)!): (n!)/(4!.(n-4)!) = 2:1 , find the valye of n. (ii) If ((2n)!)/(3!(2n-3)!): (n!)/(2!(n-2)!) = 44:3 , then find the value of n.

If n is a positive integer satisfying the equation 2+(6*2^(2)-4*2)+(6*3^(2)-4*3)+"......."+(6*n^(2)-4*n)=140 then the value of n is