Sum-up-the-following-to-nth-term-5-3-10-8-17-15-26-24-

Question Number 106515 by nimnim last updated on 05/Aug/20

S u m u p t h e f o l l o w i n g t o n t h t e r m :

\frac{5}{3} + \frac{10}{8} + \frac{17}{15} + \frac{26}{24} + \dots \dots . .

Answered by Dwaipayan Shikari last updated on 05/Aug/20

S = 1 + \frac{2}{3} + 1 + \frac{2}{8} + 1 + \frac{2}{15} + 1 + \frac{2}{24} + . .

S = (1 + 1 + 1 + \dots n) + 2 (\frac{1}{3} + \frac{1}{8} + \frac{1}{15} + \dots)

S = n + 2 \underset{n = 1}{\sum^{n}} \frac{1}{n (n + 2)}

S = n + 2 \underset{n = 1}{\sum^{n}} \frac{n + 1}{n (n + 2) (n + 1)}

S = n + 2 \sum^{n} \frac{1}{(n + 1) (n + 2)} + \frac{1}{n (n + 2) (n + 1)}

S = n + 2 \sum^{n} \frac{1}{n + 1} - \frac{1}{n + 2} + \sum^{n} \frac{n + 2 - n}{n (n + 2) (n + 1)}

S = n + 2 (\frac{1}{2} - \frac{1}{n + 2}) + \sum^{n} \frac{1}{n (n + 1)} - \sum^{n} \frac{1}{(n + 1) (n + 2)}

S = n + (1 - \frac{2}{n + 2}) + (1 - \frac{1}{n + 1}) - \sum^{n} \frac{1}{n + 1} - \frac{1}{n + 2}

S = n + (1 - \frac{2}{n + 2}) + (1 - \frac{1}{n + 1}) - \frac{1}{2} + \frac{1}{n + 2}

S = n + \frac{3}{2} - \frac{1}{n + 2} - \frac{1}{n + 1}

Commented by nimnim last updated on 06/Aug/20

T h a n k y o u S i r .

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