Find and simplify an expression for the sum of the natural numbers from (n + 1) to 2n inclusive.
i tried to solve it but couldnt i came to 0.5*(3n+1)(n-1) ...this is the answer according to my book (3/2)n^2 + n/2. could someone hint me towards getting this answer
It is an AP with the first term = ( n+1), common difference 1 and last term = 2n
Also from n+1 to 2n we have n terms.
Using expression for sum of n terms of an AP we get S = n/2 [ 2a + ( n-1) d )
Thus we have (n+1) + ( n+2) + (n+3) + ..... + 2n = n/2 [ 2(n+1) + ( n-1) x 1 ] = n/2 [ 2n + 2 + n - 1 ] = n/2[ 3n + 1 ] = (3/2 ) n^2 + (n/2)