Mathematically Proving the Solution to, $ \frac{3^2+1}{3^2-1} + \frac{5^2+1}{5^2-1} + \frac{7^2+1}{7^2-1} + ..... + \frac{99^2+1}{99^2-1} $ and Other Strategies

  TODO:
 1. Reorganize.
 2. Check for errors.

Initial Ideas:

Based on the level of expertise, this problem may be very hard or easy to solve. In this post I will write about my thinking process to solve this problem. Since this is a very long post, please report me if any mistakes are found.

By looking at the series above follows a pattern, where, the next term is an odd number. We can see the first term starts with 3, ignoring the $ +1, -1 $ for now, since they are constants in each term.

The next term can be calculated by,

So, after expanding the series above with a few more term,

$ \frac{3^2+1}{3^2-1} + \frac{5^2+1}{5^2-1} + \frac{7^2+1}{7^2-1} + \frac{9^2+1}{9^2-1} + \frac{11^2+1}{11^2-1} + ..... + \frac{97^2+1}{97^2-1} + \frac{99^2+1}{99^2-1} $

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Approximating by Calculator and Generalization:

We can easily simply the series a little further and get a rough estimate on the possible solution. Simplifying the series,

$ \ \ \ \ \frac{3^2+1}{3^2-1} + \frac{5^2+1}{5^2-1} + \frac{7^2+1}{7^2-1} + \frac{9^2+1}{9^2-1} + \frac{11^2+1}{11^2-1} + ..... + \frac{97^2+1}{97^2-1} + \frac{99^2+1}{99^2-1} $

$ = \frac{9+1}{9-1} + \frac{25+1}{25-1} + \frac{49+1}{49-1} + \frac{81+1}{81-1} + \frac{121+1}{121-1} + ..... + \frac{9409+1}{9409-1} + \frac{9801+1}{9801-1} $

$ = \frac{10}{8} + \frac{26}{24} + \frac{50}{48} + \frac{82}{80} + \frac{122}{120} + ..... + \frac{9410}{9408} + \frac{9802}{9800} $

$ \approx 1.25 + 1.083 + 1.0416 + 1.025 + 1.016 + ..... + 1.00021 + 1.0002 $

$ \approx 49.5 \ \ \ \ (\text{Here}, \ \approx \text{ means approximately}) $

Note: This is not the true answer. I will show the original answer below, first by using a computer program, then proving mathematically.

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About the calculation above:

By looking at the varying number above we can see the set is,
$ S = \{ n \ | \ n \text{ is an odd positive integer less than } 100 \} - \{1\} $

There are 100 positive integers between 1 to 100 including even and odd. So, there are half the odd numbers between 1 to 100 which is 50. The odd number set is,
$ S = \{ 1, 3, 5, 7, 11, 13, 15, .... , 95, 97, 99 \} $

We can clearly see the series above does not have 1 in it. If it did, then that could result is $ \frac{1^2+1}{1^2-1} = \frac{2}{0} $. The set can not contain 1 and we need to exclude it. So the set contains 49 positive odd integers from 3 to 99,
$ S = \{ 3, 5, 7, 11, 13, 15, .... , 95, 97, 99 \} $

Relation between set Members and Index:

$ S = \{ 1_1, 3_2, 5_3, 7_4, 11_5, 13_6, 15_7, .... , 95_{48}, 97_{49}, 99_{50} \} $

Here, the subscripts represents indices of set elements. Now, that 1 can not be a member of the set. Subtract all the indices by 1 and remove 1 element.
$ S = \{ 3_{2-1}, 5_{3-1}, 7_{4-1}, 11_{5-1}, 13_{6-1}, 15_{7-1}, .... , 95_{48-1}, 97_{49-1}, 99_{50-1} \} $

$ S = \{ 3_1, 5_2, 7_3, 11_4, 13_5, 15_6, .... , 95_{47}, 97_{48}, 99_{49} \} $

Relation between set index and the element at that index,
$ S_1 = 1 * 2 + 1 = 3 $
$ S_2 = 2 * 2 + 1 = 5 $
$ S_3 = 3 * 2 + 1 = 7 $
$...$
$...$
$...$
$ S_j = j * 2 + 1 = 2j + 1 $

So, the index of 99th term,
$2j + 1 = 99 $
$2j = 98 $
$j = 49 $

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Format of the Series:

$ = \frac{10}{8} + \frac{26}{24} + \frac{50}{48} + \frac{82}{80} + \frac{122}{120} + ..... + \frac{9410}{9408} + \frac{9802}{9800} $
Each term of the sequence is, $ \{ \frac{m+2}{m} \ | \ m \in \mathbb{Z^+} \wedge m \text{ is odd } \wedge m < 100 \} $.

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Expressing the Series as Summation:

$ \frac{3^2+1}{3^2-1} + \frac{5^2+1}{5^2-1} + \frac{7^2+1}{7^2-1} + ..... + \frac{99^2+1}{99^2-1} $

I can rewrite the series above as,
$ \frac{S_1^2+1}{S_1^2-1} + \frac{S_2^2+1}{S_2^2-1} + \frac{S_3^2+1}{S_3^2-1} + ..... + \frac{S_4^2+1}{S_4^2-1} $

Expressing it as Summation,
$ \sum\limits_{i=1}^{49} \frac{S_i^2+1}{S_i^2-1} $

or, write is as,
$ \sum\limits_{\substack{3 \leq i \leq 99 \\ i + 2}} \frac{i^2+1}{i^2-1} $

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Converting the Series Summation to Code:

I will use $ \sum\limits_{\substack{3 \leq i \leq 99 \\ i + 2}} \frac{i^2+1}{i^2-1} $ representation to convert it into code. It can be very easily translated into code. Look at the diagram below for the grouping,

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For loop starts from initial point and loops until its condition is reached and each time it increments by a certain value. Other loop types can be also implemented. Now this grouping can be translated to a loop like this,

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C++ Code:

Output of the code above:

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Patterns in the Series as Sum of Another:

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Why the last one is $ 392 $? We can calculate the value to sum by looking at the diagram above.
$ \text{value to sum} = \text{ index of the next term denominator} * 8 $.

So, we can get the next term using current value to sum. All the numerator values are multiple of $ 8 $ and the difference of two immediate numerators are also multiple of $ 8 $, which are same above and below.
$ numerator = denominator + 2 $

$ \text{next term denominator} = \text{current term denominator} + \text{value to sum} $
$ \text{next term denominator} = \text{current term denominator} + \text{index of the next term} * 8 $

So, $ \frac{7^2+1}{7^2-1} = \frac{50}{48} $ it depends on $ \frac{5^2+1}{5^2-1} = \frac{26}{24} $. $ \frac{7^2+1}{7^2-1} $ is in the $ 3rd $ position of the sequence set. Using the formula above,

$ \text{next term denominator} = 24 + 3 * 8 = 48 $ and $ numerator = denominator + 2 = 48 + 2 = 50 $
This is same as $ \frac{7^2+1}{7^2-1} = \frac{50}{48} $. Similarly, we can get the sum value for the last term denominator and numerator. We know $ \frac{99^2+1}{99^2-1} $ is in the $ 49 $ position or, index.

$ \text{value to sum} = \text{ index of the next term denominator} * 8 = 49 * 8 = 392 $. If the next term denominator is 9800 and value to sum is 392, we can get the, $ \text{previous term denominator} = 9800 - 392 = 9408 $ and the $ numerator = 9408 + 2 = 9410 $. We get 9800, when the variable n is $ 49 $, so we are trying to calculate the denominator and numerator of its precious term which has index $ 48 $.

In the $ 48th $ index, $ n = 97 $, so, $ \frac{97^2+1}{97^2-1} = \frac{9410}{9408} $. They are a match. Using the technique above we can get the any other term based on another term.

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Patterns in the Series as Multiple of Another:

Each $ denominator $ is multiple of and each $ numerator - 2 $ is also multiple of 2. There are 49 terms in the series. So, there are 48 plus signs in between. Ex: We can partition two things with one symbol, $ \Box \ | \ \Box $. Similarly, we can partition two things with one symbol, $ \Box \ | \ \Box | \ \Box $. so, if there are $ m $ things we need $ m - 1 $ symbols to partition.

There are 48 multiples of 8 to generate the next terms from the first term. Each time they are incremented by 1 starting from 3. So the progression is,
$ 3 + 4 + 5 + 6 + 7 + ..... + 49 $

We know sum from 1 to n is, $ \sum_{i=1}^{n} i = \frac{n*(n+1)}{2} $. Get the multiple of 8 denominator using the index in the set.
$ denominator = \frac{index*(index+1)}{2} * 8 $. For, the last one( 49th term ) $ index = 49 $, $ denominator = \frac{49*(49+1)}{2} * 8 = 1225 * 8 = 9800 $. We can also get the previous term multiple by, $ \text{next term multiple} = \text{sum to get next term} + ( \text{current term denominator} / 8 ) $.

The previous term is $ \frac{97^2+1}{97^2-1} $, where, $ n = 48 $.
$ ( \text{next term multiple} - \text{sum to get next term} ) * 8 = \text{current term denominator} $
$ \text{current term denominator} = ( 1225 - 49 ) * 8 = 9408 $. Also the denominator of $ \frac{97^2+1}{97^2-1} $ is $ \frac{9410}{9408} $.

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Pattern in the series $ 1 + \frac{1}{3} + \frac{1}{6} + \frac{1}{10} + \frac{1}{15} + \frac{1}{21} + ..... + \frac{1}{1225}: $

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Solving the Series $ 1 + \frac{1}{3} + \frac{1}{6} + \frac{1}{10} + \frac{1}{15} + \frac{1}{21} + ..... + \frac{1}{1225}: $

So far we have,
$ 49 + \frac{1}{4} * \Bigg( 1 + \frac{1}{3} + \frac{1}{6} + \frac{1}{10} + \frac{1}{15} + \frac{1}{21} + ..... + \frac{1}{1225} \Bigg) $
Since the true answer is 49.49, we can deduce that,
$ \frac{1}{4} * \Bigg( 1 + \frac{1}{3} + \frac{1}{6} + \frac{1}{10} + \frac{1}{15} + \frac{1}{21} + ..... + \frac{1}{1225} \Bigg) = 0.49 $
So, the sequence above should result in,
$ 1 + \frac{1}{3} + \frac{1}{6} + \frac{1}{10} + \frac{1}{15} + \frac{1}{21} + ..... + \frac{1}{1225} = 1.96 $
At first glance this looks very hard, but it too can be calculated mathematically. The Sequence set of the above series is,
$ P = \{ 1, \frac{1}{3}, \frac{1}{6}, \frac{1}{10}, \frac{1}{15}, \frac{1}{21}, ....., \frac{1}{1225} \} $
Relation between set index and elements,
$ P = \{ P_1, P_2, P_3, P_4, P_5, P_6, ....., P_{49} \} $
$ P_1 = \frac{2}{1*(1+1)} = 1 $

$ P_2 = \frac{2}{2*(2+1)} = \frac{1}{3} $

$ P_3 = \frac{2}{3*(3+1)} = \frac{1}{6} $

$ P_4 = \frac{2}{4*(4+1)} = \frac{1}{10} $

$ P_5 = \frac{2}{5*(5+1)} = \frac{1}{15} $

$ P_6 = \frac{2}{6*(6+1)} = \frac{1}{21} $
$ ... $
$ ... $
$ ... $
$ P_{49} = \frac{2}{49*(49+1)} = \frac{1}{1225} $

So, we can calculate the $ i $ th member of $ P $ set using formula,
$ P_i = \frac{2}{i*(i+1)} $

The summation can be expressed as,
$ S_P = \sum_{i=1}^{49} \frac{2}{i*(i+1)} $



The problem still isn't solved yet, since it is summation from and still needs to be calculated. Now express the terms in summation in such a way that some terms are canceled and we are left with a simple form.
We can represent $ \frac{2}{i*(i+1)} $ as,
$ \frac{2}{i*(i+1)} \equiv \frac{A}{i} + \frac{B}{i+1} $
Now, multiply $ A $ with $ (i+1) $ and $ B $ with $ Bi $ which equals 2, since our numerator have be 2 to keep them equivalent.
$ 2 = A*(i+1) + B*i $

To calculate value of $ A $ and $ B $, choose i in a way such that the other one becomes zero,
if, $ i = 0 $, $ 2 = A*(0+1) + B*0, So, A = 2 $

if, $ i = -1 $, $ 2 = A*(-1+1) + B*(-1), So, B = -2 $

Now we can write the original equation as, $ \frac{2}{i*(i+1)} = \frac{2}{i} + \frac{-2}{i+1} $

$ \frac{2}{i*(i+1)} = 2 * \Bigg( \frac{1}{i} - \frac{1}{i+1} \Bigg) $

The sum of the sequence from $ i = 1 \text{ to } n $ is,

$ \sum_{i=1}^{n} \frac{2}{i*(i*1)} = \sum_{i=1}^{n} \frac{2}{i} + \frac{-2}{i+1} $

$ \sum_{i=1}^{n} \frac{2}{i*(i*1)} = 2 * \sum_{i=1}^{n} \Bigg( \frac{1}{i} - \frac{1}{i+1} \Bigg) $

Here, $ n = 49 $ since we need to capitulate the sum of 49 terms,

$ \sum_{i=1}^{49} \Bigg( \frac{1}{i} - \frac{1}{i+1} \Bigg) $

Expanding the summation,

$ \require{cancel} 2 * \sum_{i=1}^{n} \Bigg( \frac{1}{i} - \frac{1}{i+1} \Bigg) = 2 * [ ( \frac{1}{1} - \cancel{\frac{1}{2}} ) + ( \cancel{\frac{1}{2}} - \cancel{\frac{1}{3}} ) + ( \cancel{\frac{1}{3}} - \cancel{\frac{1}{4}} ) + ( \cancel{\frac{1}{4}} - \cancel{\frac{1}{5}} ) + ..... + ( \cancel{\frac{1}{n-1}} - \cancel{\frac{1}{n}} ) + ( \cancel{\frac{1}{n}} - \frac{1}{n+1} ) ] $

$ 2 * \sum_{i=1}^{n} \Bigg( \frac{1}{i} - \frac{1}{i+1} \Bigg) = 2 * [ 1 - \frac{1}{1+n} ] $

So, $ \sum_{i=1}^{n} \frac{2}{i*(i*1)} = 2 * \sum_{i=1}^{n} \Bigg( \frac{1}{i} - \frac{1}{i+1} \Bigg) = \frac{2n+2-2}{1+n} = \frac{2n}{1+n} $

Recall that, $ \sum_{i=1}^{n} \frac{2}{i*(i*1)} $ represents sum of series $ 1 + \frac{1}{3} + \frac{1}{6} + \frac{1}{10} + \frac{1}{15} + \frac{1}{21} + ..... + \frac{1}{1225} $, where, $ n = 49 $.

$ 1 + \frac{1}{3} + \frac{1}{6} + \frac{1}{10} + \frac{1}{15} + \frac{1}{21} + ..... + \frac{1}{1225} = \frac{2n}{1+n}, \text{ where, n = 49} $

$ 1 + \frac{1}{3} + \frac{1}{6} + \frac{1}{10} + \frac{1}{15} + \frac{1}{21} + ..... + \frac{1}{1225} = \frac{2*49}{1+49} $

$ 1 + \frac{1}{3} + \frac{1}{6} + \frac{1}{10} + \frac{1}{15} + \frac{1}{21} + ..... + \frac{1}{1225} = \frac{98}{50} $

$ 1 + \frac{1}{3} + \frac{1}{6} + \frac{1}{10} + \frac{1}{15} + \frac{1}{21} + ..... + \frac{1}{1225} = 1.96 \ \ \ \ \Box $

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Results from Wolfram Alpha:

Input:

Alternatively, follow the link below to get the result.
https://www.wolframalpha.com/input/?i=Sum%5B(k%5E2%2B1)%2F(k%5E2-1),%7Bk,3,99,2%7D%5D

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Solution to the Problem:

$ \ \ \ \ \frac{3^2+1}{3^2-1} + \frac{5^2+1}{5^2-1} + \frac{7^2+1}{7^2-1} + \frac{9^2+1}{9^2-1} + ..... + \frac{99^2+1}{99^2-1} $

$ = \frac{10}{8} + \frac{26}{24} + \frac{50}{48} + \frac{82}{80} + ..... + \frac{9802}{9800} $

$ = \frac{8+2}{8} + \frac{24+2}{24} + \frac{48+2}{48} + \frac{80+2}{80} + ..... + \frac{9800+2}{9800} $

$ = \frac{8}{8} + \frac{2}{8} + \frac{24}{24} + \frac{2}{24} + \frac{48}{48} + \frac{2}{48} + \frac{80}{80} + \frac{2}{80} + ..... + \frac{9800}{9800} + \frac{2}{9800} $

$ = 1 + \frac{2}{8} + 1 + \frac{2}{24} + 1 + \frac{2}{48} + 1 + \frac{2}{80} + ..... + 1 + \frac{2}{9800} $

$ = 49 + \frac{2}{8} + \frac{2}{24} + \frac{2}{48} + \frac{2}{80} + ..... + \frac{2}{9800} $

$ = 49 + \frac{1}{4} + \frac{1}{12} + \frac{1}{24} + \frac{1}{40} + ..... + \frac{1}{4900} $

$ = 49 + \frac{1}{4} * \Bigg( 1 + \frac{1}{3} + \frac{1}{6} + \frac{1}{10} + \frac{1}{15} + \frac{1}{21} + ..... + \frac{1}{1225} \Bigg) $

$ = 49 + \frac{1}{4} * \frac{2*49}{49+1} $

$ = 49 + \frac{1}{4} * 1.96 $

$ = 49.49 \ \ \ \ \Box $