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Question

If N denotes set of all positive integers and if f:NN is defined by f(n) = the sum of positive divisors of n, then f(2k.3) where k is a positive integer is

A
2k+11
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B
2(2k+11)
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C
3(2k+11)
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D
4(2k+11)
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Solution

The correct option is D 4(2k+11)
The divisors of (2k)(3) are 1,2,22,.......,2k,3,3×2,3×22,.......,3×2k
If you think about it, the sum of all thepositiv divisors of (2k)(3) can be conveniently written as:
(20+21+22+...+2k)(30+31)
Now, 20,21,22,........,2k forms a geometric series.
Where a=20=1,r=2
Using the formula for the sum of a GP:
Sn=a(rn1)(r1)
=1(2k+11)(21)×(1+3)
=4(2k+11)

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