An algebra problem by Tan T

Algebra Level 3

3 1 × 2 ( 1 2 ) + 4 2 × 3 ( 1 4 ) + 5 3 × 4 ( 1 8 ) + ⋯ \dfrac 3{1 \times 2} \left( \dfrac 12 \right) + \dfrac 4{2 \times 3} \left( \dfrac 14 \right) + \dfrac 5{3 \times 4} \left( \dfrac 18 \right) + \cdots

Find the sum of the first 9 terms of the summation above.

511 512 \dfrac{511}{512} 10239 10240 \dfrac{10239}{10240} 5119 5120 \dfrac{5119}{5120} 1023 1024 \dfrac{1023}{1024}

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Tan T by Tan T , 27, India

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1 solution

Chew-Seong Cheong
May 26, 2016

S = 3 1 × 2 ( 1 2 ) + 4 2 × 3 ( 1 4 ) + 5 3 × 4 ( 1 8 ) + ⋯ + 11 9 × 10 ( 1 2 9 ) = ∑ n = 1 9 n + 2 2 n n ( n + 1 ) = ∑ n = 1 9 1 2 n ( 1 n + 1 + 2 n ( n + 1 ) ) = ∑ n = 1 9 1 2 n ( 1 n + 1 + 2 n − 2 n + 1 ) = ∑ n = 1 9 1 2 n ( 2 n − 1 n + 1 ) = ∑ n = 1 9 ( 1 2 n − 1 n − 1 2 n ( n + 1 ) ) = ∑ n = 1 9 1 2 n − 1 n − ∑ n = 2 10 1 2 n − 1 n = 1 2 0 ⋅ 1 − 1 2 9 ⋅ 10 = 1 − 1 5120 = 5119 5120 \begin{aligned} S & = \frac 3{1 \times 2} \left( \frac 12 \right) + \frac 4{2 \times 3} \left( \frac 14 \right) + \frac 5{3 \times 4} \left( \frac 18 \right) + \cdots + \frac {11}{9 \times 10} \left( \frac 1{2^9} \right) \\ & = \sum_{n=1}^9 \frac{n+2}{2^nn(n+1)} \\ & = \sum_{n=1}^9 \frac 1 {2^n}\left(\frac 1{n+1} + \frac 2{n(n+1)} \right) \\ & = \sum_{n=1}^9 \frac 1 {2^n}\left(\frac 1{n+1} + \frac 2n - \frac 2{n+1} \right) \\ & = \sum_{n=1}^9 \frac 1 {2^n}\left(\frac 2n - \frac 1{n+1} \right) \\ & = \sum_{n=1}^9 \left( \frac 1{2^{n-1}n} - \frac 1{2^n(n+1)} \right) \\ & = \sum_{n=1}^9 \frac 1{2^{n-1}n} - \sum_{n=2}^{10} \frac 1{2^{n-1}n} \\ & = \frac 1{2^0\cdot 1} - \frac 1{2^9\cdot 10} \\ & = 1 - \frac 1{5120} \\ & = \boxed{\dfrac{5119}{5120}} \end{aligned}

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thumbs up!

Tan T - 5 years ago

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Thanks, shouldn't you upvote it?. I edited your problem. Hope you learn up LaTex soon. You can see the keystrokes by placing mouse cursors on the formulas.

Chew-Seong Cheong - 5 years ago

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