Tough sigma

Algebra Level 3

∑ n = 1 99 5 100 2 5 n + 5 100 = ? \large \sum_{n=1}^{99} \dfrac{5^{100}}{25^n + 5^{100}} = \ ?


The answer is 49.5.

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Harsh Poddar by harsh poddar , 22, India

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2 solutions

Ben Habeahan
Aug 19, 2015

∑ n = 1 99 5 100 2 5 n + 5 100 = ∑ n = 1 99 2 5 50 2 5 n + 2 5 50 = ∑ n = 1 99 1 2 5 n − 50 + 1 = ∑ n = 1 100 1 2 5 n − 50 + 1 − ∑ n = 100 100 1 2 5 n − 50 + 1 = 1 2 5 − 49 + 1 + 1 2 5 − 48 + 1 + 1 2 5 − 47 + 1 + ⋯ + 1 2 5 − 1 + 1 + 1 2 5 0 + 1 + 1 2 5 1 + 1 + ⋯ + 1 2 5 47 + 1 + 1 2 5 48 + 1 + 1 2 5 49 + 1 + 1 2 5 50 + 1 − 1 2 5 50 + 1 ( ∗ ) \displaystyle \sum_{n=1}^{99} \frac{5^{100}}{25^{n}+5^{100}} = \displaystyle \sum_{n=1}^{99} \frac{25^{50}}{25^{n}+25^{50}} \\ = \displaystyle \sum_{n=1}^{99} \frac{1}{25^{n-50}+1} {\\} = \displaystyle \sum_{n=1}^{100} \frac{1}{25^{n-50}+1} - \displaystyle \sum_{n=100}^{100}\frac{1}{25^{n-50}+1} \\ = \frac{1}{25^{-49}+1}+\frac{1}{25^{-48}+1}+ \frac{1}{25^{-47}+1}{\\} + \dots+ \frac{1}{25^{-1}+1} + \frac{1}{25^{0}+1} +\frac{1}{25^{1}+1}{\\}+ \dots + \frac{1}{25^{47}+1}+ \frac{1}{25^{48}+1}+ \frac{1}{25^{49}+1} {\\}+ \frac{1}{25^{50}+1}-\frac{1}{25^{50}+1} (*) \\ use fact, 1 2 5 k + 1 + 1 2 5 − k + 1 = 2 5 − k + 1 + 2 5 k + 1 2 5 − k + k + 2 5 − k + 2 5 k + 1 = 1 \\ \frac{1}{25^{k}+1} + \frac{1}{25^{-k}+1} = \frac{25^{-k}+1+25^{k}+1}{25^{-k+k}+25^{-k}+25^{k}+1} =1 \\ so the result (*) will be 49 × 1 + 1 2 = 49.5 49 \times 1 + \frac{1}{2} = \boxed {49.5}

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If anyone is confused, the extra 1/2 comes from 1 2 5 0 + 1 \frac{1}{25^0+1} term. Nice solution!

Scott Ripperda - 5 years, 9 months ago 
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for n=1 and n=99:

5**100 / (5**2 + 5**100) + 5**100 / (5**198 + 5**100)


5**98 / (1 + 5**98) + 1 / (5**98 + 1)
(5**98 + 1) / (5**98 + 1) = 1

There are 49 pairs
1*49 = 49

when n=50: (middle term)
the result will be 1/2

Answer: 49 + 1/2 --> 49

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