Find the sum of the series
11⋅3+13⋅5+15⋅7+17⋅9+19⋅11+⋯
My attempt solution:
13⋅(1+15)+17⋅(15+19)+111⋅(19+113)+⋯
=13⋅(65)+17⋅(1445)+111⋅(22117)+⋯
=2⋅((15)+(145)+(1117)+⋯)
=2⋅((15)+(15⋅9)+(19⋅13)+⋯)
It is here that I am stuck. The answer should be 12 but I don't see how to get it. Any suggestions?
Also, a bit more generally, are there good books (preferably with solutions) to sharpen my series skills?
Answer
This is a general approach to evaluate the sum of series, like these.
First find nth term of series.
Let Tn denote the nth term.
We see that,
T1=11⋅3
T2=13⋅5
And so on.
Let the numbers in green be
X1,X2,X3,X4,..=1,3,5,7...
Clearly they form an A.P. with common difference =2
So, nth term of this AP is 1+(n−1)×2=2n−1
Similarly,
Let the numbers in teal be Y1,Y2,Y3,Y4,..=3,5,7,9...
Clearly they form an A.P. with common difference =2
So, nth term of this AP is 3+(n−1)×2=2n+1
So, the nth term of the main question is just
Tn=1(2n−1)⋅(2n+1)
Now, taking summation from 1 to n , we have,
n∑n=11(2n−1)⋅(2n+1)
=n∑n=1=12⋅(2n+1)−(2n−1)(2n−1)(2n+1)
=n∑n=1=12⋅1(2n−1)−1(2n+1)
=n∑n=1=12⋅(1−1(2n+1))
While n=∞,
∞∑n=1=12⋅(1−1(2(∞)+1))
=∞∑n=1=12⋅(1−1∞)
Since 1∞=0,
∞∑n=1=12⋅(1−0)
Which is
∞∑n=1=12
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