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For all positive integers n and m, the function A(n) equals

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For all positive integers n and m, the function A(n) equals

by AAPL » Tue May 29, 2018 6:54 am

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For all positive integers n and m, the function A(n) equals the following product:
(1 + 1/2 + 1/2^2)(1 + 1/3 + 3^2)(1 + 1/5 + 5^2)...(1 + 1/p_n + 1/p_n^2), where p_n is the nth smallest prime number, while B(m) equals the sum of the reciprocals of all the positive integers from 1 through m, inclusive. The largest reciprocal of an integer in the sum that B(25) represents that is NOT present in the distributed expansion of A(5) is,

A. 1/4
B. 1/5
C. 1/6
D. 1/7
E. 1/8

The OA is E.

Is there a strategic approach to solve this question? Can anyone help, please? Thanks!
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answer

by Vincen » Fri Jun 01, 2018 12:21 am
Hi AAPL.

First of all, I think there is a mistake in the formula:
A(n) equals the following product: (1 + 1/2 + 1/2^2)(1 + 1/3 + 3^2)(1 + 1/5 + 5^2)...(1 + 1/p_n + 1/p_n^2)
It should be (1 + 1/2 + 1/2^2)(1 + 1/3 + 1/3^2)(1 + 1/5 + 1/5^2)...(1 + 1/p_n + 1/p_n^2).

Now, first let's write B(25). $$B\left(25\right)=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}+\frac{1}{9}+\cdots+\frac{1}{24}+\frac{1}{25}.$$ Now, let's write A(5). But, we have to remember that 5 represents the fifth smalles prime number, hence this sum involves the prime numbers: {2, 3, 5, 7, 11}.

Hence, we get $$A\left(5\right)=\left(1+\frac{1}{2}+\frac{1}{2^2}\right)\left(1+\frac{1}{3}+\frac{1}{3^2}\right)\left(1+\frac{1}{5}+\frac{1}{5^2}\right)\left(1+\frac{1}{7}+\frac{1}{7^2}\right)\left(1+\frac{1}{11}+\frac{1}{11^2}\right).$$

We have to look from left to right B(25) and find the first reciprocal that doesn't appear on A(5). But first, if we apply the distributivity property and reorder a little bit the terms in A(5) we can write it as follows: $$A\left(5\right)=\left(1+\frac{1}{2}+\frac{1}{2^2}\right)\left(1+\frac{1}{3}+\frac{1}{3^2}\right)\left(1+\frac{1}{5}+\frac{1}{5^2}\right)\left(1+\frac{1}{7}+\frac{1}{7^2}\right)\left(1+\frac{1}{11}+\frac{1}{11^2}\right)$$ $$=\left(1+\frac{1}{2}+\frac{1}{4}\right)\left(1+\frac{1}{3}+\frac{1}{9}\right)\left(1+\frac{1}{5}+\frac{1}{25}\right)\left(1+\frac{1}{7}+\frac{1}{49}\right)\left(1+\frac{1}{11}+\frac{1}{121}\right)$$ $$=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\left(\frac{1}{2}\cdot\frac{1}{3}\right)+\frac{1}{7}+\frac{1}{9}+\left(\frac{1}{2}\cdot\frac{1}{5}\right)+\frac{1}{11}+\left(\frac{1}{4}\cdot\frac{1}{3}\right)+\left(\frac{1}{2}\cdot\frac{1}{7}\right)+\cdots$$ $$=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{9}+\frac{1}{10}+\frac{1}{11}+\frac{1}{12}+\frac{1}{14}+\cdots$$ If we look carefully, 1/8 is the first term of B(25) that doesn't appear on A(5). The second one is 1/13.

Hence, the largest reciprocal of an integer in the sum that B(25) represents that is NOT present in the distributed expansion of A(5) is 1/8.

Therefore, the correct answer is the option E.

I hope it helps you.

Regards.
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