Easy Summation

Algebra Level 3

∑ i = 1 48 log ⁡ 17 ( i + 3 i + 2 ) = ? \sum_{i=1}^{48} \log_{17} \left( \dfrac{i+3}{i+2} \right) = \, ?

ln(3/4e) infinity 0 ln(4/3e) 1 ln(3e/4) ln(4e/3)

Prince Loomba by Prince Loomba , 21, India

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2 solutions

∑ i = 1 48 log ⁡ 17 ( i + 3 i + 2 ) = log ⁡ 17 ( ∏ i = 1 48 i + 3 i + 2 ) \displaystyle \sum_{i=1}^{48} \log_{17}(\frac{i+3}{i+2})=\log_{17}(\prod_{i=1}^{48} \frac{i+3}{i+2}) log ⁡ 17 ( ∏ i = 1 48 ( ( i + 1 ) + 2 i + 2 ) \log_{17}(\prod_{i=1}^{48}(\frac{(i+1)+2}{i+2}) log ⁡ 17 ( 51 3 ) = log ⁡ 17 ( 17 ) = 1 \log_{17}(\frac{51}{3})=\log_{17}(17)=1

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Prince Loomba
Jan 26, 2016

ln4-ln3+ln5-ln4+....+ln51-ln50=ln51-ln3=ln17=1 here i have used ln for log to base 17

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