Find the sum of the first n terms of the series $1^3+3(2)^2+3^3+3(4)^2+5^3\dots$ If (i) n is even and (ii)if n is odd? I tried regrouping and finding the sums of the cubic series and the even squared one but it's not working.Should I regroup or find the general term?
A progression and series question.
0
$\begingroup$
summation
-
0Usually, this is where you tell us what the general term of the series is. – 2017-02-10
-
0What is 3(2)^2? Is it $3\cdot 2^2$ or is it $(3\cdot 2)^2$? What is the general term in the series? – 2017-02-10
-
1$1+3(2^2+3^2+4^2+...)$ – 2017-02-10
-
0you should confirm that the next term is $3(5^2)$ and not $5^3$... – 2017-02-10
-
0The $1^3$ looks out of place. The series could very well be $$1^3+3(2^2)+3^3+3(4^2)+5^3+3(6^2)+7^3+3(8^2)+\cdots$$ – 2017-02-10
-
0Yes it is! I should have elaborated – 2017-02-10
1 Answers
1
We can write our sum as: $$ S = 1^3 +3 (2^2 +3^2 +\cdots n^2) $$ $$S =1 +3 (\sum_{i=2}^{n} i^2) $$ $$S = 1+ 3 (\sum_{i=1}^{n} i^2 -1) $$
Hope you can take it from here.
-
2And the sum of squares: http://math.stackexchange.com/a/2015855/272831 – 2017-02-10
-
0Sorry,first time,and the next term is 5^3 – 2017-02-10