Series Demystified 6

Algebra Level 4

$\large \sum_{n=1}^{48} \log \left ( \frac {n}{n^2+3n+2} \right )$

If the summation above equals to $S$ , then $S+\log(49!)=-\log P.$ Find the value of $P$ .

This is one of my original Series Demystified problems .

The answer is 1225.

2 solutions

Vishwak Srinivasan
Jun 20, 2015

$\log\dfrac{n}{n^2 + 3n + 2} = \log\dfrac{n}{(n+1)(n+2)} = \log(n) - \log(n+1) - \log(n+2)$

$\displaystyle \sum_{n=1}^{48} \log(n) = \log(1) + \log(2) + \ldots + \log(48) = \log(48!)$

$\displaystyle \sum_{n=1}^{48} \log(n+1) = \log(2) + \log(3) + \ldots + \log(49) = \log(1) + \log(2) + \log(3) + \ldots + \log(49) = \log(49!)$

$\displaystyle \sum_{n=1}^{48} \log(n+2) = \log(3) + \log(4) + \ldots + \log(50) = \log(50!) - \log(2) - \log(1) = \log(50!) - \log(2)$

$S = \log(48!) - \log(49!) - \log(50!) + \log(2)$

$S + \log(49!) = \log(\dfrac{48!}{50!}) + \log(2)$

$S + \log(49!) = -\log(49.50) + \log(2)$

$S + \log(49!) = -\log(\dfrac{49.50}{2})$

$P = \dfrac{49.50}{2} =1225$

Did the same

Aditya Kumar - 5 years, 1 month ago

I really love this set!!!

I Gede Arya Raditya Parameswara - 4 years, 3 months ago

I did almost the same way.

Niranjan Khanderia - 2 years, 11 months ago

Chew-Seong Cheong
Jun 20, 2015

$\begin{aligned} S & = \sum_{n=1}^{48} {\log{\frac{n}{(n^2+3n+2)}}} = \sum_{n=1}^{48} {\log{\frac{n}{(n+1)(n+2)}}} \\ & = \log{\frac{48!\dot{} 2}{49!50!}} = \log{\frac{2}{49\dot{} 50!}} = \log{2} - \log{49} - \log{(50!)} \end{aligned}$

$\begin{aligned} S + \log{(49!)} & = \log{2} - \log{49} - \log{(50!)} + \log{(49!)} \\ & = \log{2} - \log{49} - \log{50} - \log{(49!)} + \log{(49!)} \\ & = - \log{(49\dot{}25)} \\ & = - \log{1225} \\ \Rightarrow P & = \boxed{1225} \end{aligned}$

Moderator note:

That's a nice way to solve this without dealing with telescoping sum.

Nice solution sir :)

Refaat M. Sayed - 5 years, 11 months ago

Without dealing with telescoping sum. (+1