Factorise the expression: am^2 + bm^2 + bn^2+an^2
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The series converges to 1/(1-9x) for -1/9<x<1/9
Given the series is ∑
We have to find the values of x for which the series converges.
We know,
∑ converges to (a) / (1-r) if r < 1
Otherwise the series will diverge.
Here, ∑ is a geometric series with |r| = | 9x |
And it converges for |9x| < 1
Hence, the given series gets converge for -1/9<x<1/9
And geometric series converges to a/(1-r)
Here, a = 1 and r = 9x
Therefore, a/(1-r) = 1/(1-9x)
Hence, the given series converges to 1/1-9x for -1/9<x<1/9
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