Calculate the sum of n terms of the string An = n^2 + 12n.
2 Answers
justaguide | College Teacher | (Level 2) Distinguished Educator
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
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]