Prove that square of any even natural number i.e. `(2n)^2` is equal to sum of n terms of a certain series of integers in `A.P.`

Prove that (2n)^2 where n in N can be expressed as the sum of n terms of a series of integers in AP

The sum of n terms of a series is (2n^(2)+n) .Show that it is an A.P. .

The sum of n terms of a series is (3n^(2)+2n) . Prove that it is an A.P.

If the sum of m terms of an A.P, is equal to the sum of n terms of the A.P ., then the sum of (m + n) terms is

The sum of n terms of an A.P. is 4n^(2) + 5n. Find the series.

The sum of first n terms of a certain series is given as 2n^(2)-3n . Show that the series is an A.P.

Prove that the sum of any positive integer and its square is an even number.

Assertion (A) : Sum of first hundred even natural numbers divisible by 5 is 500 . Reason (R ) : Sum of first n terms of an A.P is given by S_(N)=(n)/(2)[a+l],l is last term

Statement - I : The variance of first n even natural numbers is (n^(2) - 1)/(4) Statement - II : The sum of first n natural numbers is (n(n+1))/(2) and the sum of the squares of first n natural numbers is (n(n+1)(2n+1))/(6)

If in an A.P. the sum of m terms is equal to n and the sum of n terms is equal to m then prove that the sum of (m+n) terms is -(m+n)