# Calculate the sum of n terms of the string An = n^2 + 12n.

justaguide | College Teacher | (Level 2) Distinguished Educator

Posted on

We have to find the sum of the terms An = n^2 + 12n

Now Sum [ An], of n terms

=> Sum [ n^2 + 12n ] , of n terms

=> Sum [ n^2] + Sum [ 12n] , of n terms

=> n(n+1)(2n+1)/6 + 12* n(n+1)/2

=> n(n+1)(2n+1)/6 + 6*n(n+1)

=> n(n+1) [ (2n+1)/6 + 6]

=> n(n+1)[ (2n + 1 + 36)/6)]

=> n(n+1)(2n + 37)/ 6

=> (1/6)*n*(n+1)*(2n+ 37)

The required sum is (1/6)*n*(n+1)*(2n+ 37)

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

We'll re-write the sum of the terms of the given string, whose general terms is  an = n^2 + 12n:

Sum (k^2 + 12k)

Sum (k^2 + 12k) = Sum k^2 + Sum 12k

Sum k^2 = 1^2 + 2^2 + ... + n^2 = n(n+1)(2n+1)/6 (1)

Sum 12k= 12*Sum k

Sum k = 1 + 2 + 3 + .... + n

The sum of the first n natural terms is:

Sum k = n(n+1)/2

12*Sum k = 12n*(n +1)/2

We'll simplify and we'll get:

12*Sum k = 6n*(n +1)(2)

Sum (k^2 + 12k) = (1) + (2)

Sum (k^2 + 12k) = n(n+1)(2n+1)/6 + 6n*(n +1)

We'll factorize by n(n+1):

Sum (k^2 + 12k) = [n(n+1)]*[(2n+1)/6+ 6]

Sum (k^2 + 12k) = [n(n+1)]*[(2n + 1 + 36)/6]

Sum (k^2 + 12k) = [n(n+1)]*[(2n + 37)/6]

So, the value of the sum of the terms of the string is:

Sum (k^2 + 12k) = [n(n+1)]*[(2n + 37)/6]