Complicated Sum

Algebra Level 3

A = ∑ k = 1 100 1 ( ∑ a = 0 ∞ 1 2 a ) 2 + 1 ∑ b = 0 k + 1 ( 2 b + 1 ) + 1 ∑ c = 0 k − 1 ( 2 c + 1 ) \large A=\sum_{k=1}^{100}\sqrt{\frac{1}{\left(\sum_{a=0}^{\infty}\frac{1}{2^{a}}\right)^2}+\frac{1}{\sum_{b=0}^{k+1}(2b+1)}+\frac{1}{\sum_{c=0}^{k-1}(2c+1)}}

Given the above, find 10302 A 10302A .


The answer is 530350.

Chan Tin Ping by Chan Tin Ping , 20, Malaysia

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1 solution

Chan Tin Ping
Oct 22, 2017

First, separate the summation. i)(geometric progression) ∑ a = 0 ∞ 1 2 a = 2 \sum_{a=0}^{\infty}\frac{1}{2^{a}}=2 ii)(arithmetic progression) ∑ b = 0 k + 1 ( 2 b + 1 ) = ( k + 2 ) 2 \sum_{b=0}^{k+1}(2b+1)=(k+2)^2 iii) ∑ c = 0 k − 1 ( 2 c + 1 ) = k 2 \sum_{c=0}^{k-1}(2c+1)=k^2

A = ∑ k = 1 100 1 2 2 + 1 ( k + 2 ) 2 + 1 k 2 = ∑ k = 1 100 ( 1 2 ) 2 + ( 1 k − 1 k + 2 ) 2 + 2 k ( k + 2 ) = ∑ k = 1 100 ( 1 2 ) 2 + ( 1 k − 1 k + 2 ) 2 + ( 1 k − 1 k + 2 ) = ∑ k = 1 100 ( 1 2 + 1 k − 1 k + 2 ) 2 = ∑ k = 1 100 1 2 + ∑ k = 1 100 ( 1 k − 1 k + 2 ) = 50 + 1 + 1 2 − 1 101 − 1 102 \begin{aligned} A&=\sum_{k=1}^{100}\sqrt{\frac{1}{2^2}+\frac{1}{(k+2)^2}+\frac{1}{k^2}}\\ &=\sum_{k=1}^{100}\sqrt{(\frac{1}{2})^2+(\frac{1}{k}-\frac{1}{k+2})^2+\frac{2}{k(k+2)}}\\ &=\sum_{k=1}^{100}\sqrt{(\frac{1}{2})^2+(\frac{1}{k}-\frac{1}{k+2})^2+(\frac{1}{k}-\frac{1}{k+2})}\\ &=\sum_{k=1}^{100}\sqrt{(\frac{1}{2}+\frac{1}{k}-\frac{1}{k+2})^2}\\ &=\sum_{k=1}^{100}\frac{1}{2}+\sum_{k=1}^{100}(\frac{1}{k}-\frac{1}{k+2})\\ &=50+1+\frac{1}{2}-\frac{1}{101}-\frac{1}{102} \end{aligned} Hence, 10302 A = 530350 10302A=530350

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