Inspired By Shubhendra Singh

Probability Level 5

1 2 3 × 3 ! − 1 × 3 2 4 × 4 ! + 1 × 3 × 5 2 5 × 5 ! − … \large \dfrac{1}{2^3 \times3!}-\dfrac{1\times 3}{2^4\times 4!}+\dfrac{1\times3\times5}{2^5\times5!} - \ldots

If the alternating series above equals to A 24 − B B + 1 2 \dfrac A{24} - \dfrac B{B+1} \sqrt2 for positive integers A A and B B , find the value of A + B A+B .


The answer is 25.

Kritarth Lohomi by kritarth lohomi , 18, India

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1 solution

Tom Van Lier
Jul 21, 2016

We will Taylor sqrt{(1+x)^3 around x = 0 :

( 1 + x ) 3 = 1 + 3 × x 2 + 3 × x 2 2 2 × 2 ! − 3 × x 3 2 3 × 3 ! + 3 × 3 × x 4 2 4 × 4 ! − 3 × 3 × 5 × x 5 2 5 × 5 ! + . . . \sqrt{(1+x)^3} = 1 + \dfrac{3 \times x}{2} + \dfrac{3 \times x^2}{2^2 \times 2!} - \dfrac{3 \times x^3}{2^3 \times 3!} + \dfrac{3 \times 3 \times x^4}{2^4 \times 4!} - \dfrac{3 \times 3 \times 5 \times x^5}{2^5 \times 5!} + ... .

With the notation that S S is the given sum, and plugging in 1 for x , we get :

( 2 ) 3 = 23 8 − 3 × S \sqrt{(2)^3 }= \dfrac{23}{8} - 3 \times S .

Solving for S, we get S = 23 24 − 2 2 3 S = \dfrac{23}{24} - \dfrac{2 \sqrt{2}}{3} , such that A + B = 25 A + B = 25 .

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