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# Algebra

Find the sum of infinite numbers of the set S ∈ {1,1/2,1/3,1/4,1/6,1/8,1/9,1/12,1/16,..}. It is given that each term in S is of the form 1/(2a ×3b ) where a, b are non-negative integers.

1. 2
2. 3
3. 4
4. infinite

### 5 comments

1. Answer is 3.

The given series is a product of two infinite series: (1+1/2+1/4+…..).(1+1/3+1/9+1/81+. .)= 2.(3/2)=3

1. Option 2.

2. Abhishek Tandon

3

Approach:
The above series is a product of 2 infinte GP.

Also,
1/2^a*3^b can be expanded as
1/2^0*3^0 + 1/2^1*3^0 + 1/2^0*3^1 + 1/2^1*3^1 +….
Which is nothing but equal to
1+1/2+1/3+1/6+….
as a and b values are increased the sequence will increase.

The above series is actually a product of two gp’s
(1+1/2)(1+1/3) = 1+1/2+1/3+1/6…. Which is the same equation.
Hence we can find
GP sum = a/1-r
2*3/2=3

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