Ancient Sum

Algebra Level 4

∑ n = 1 25 2 n − 1 n ( n + 1 ) ( n + 2 ) \large \displaystyle\sum_{n=1}^{25}\frac{2n-1}{n(n+1)(n+2)}

If the summation above equals to M E \frac ME for coprime positive coprime integer, then find the value of M + E M+E .


The answer is 1177.

View wiki
Shivam Jadhav by Shivam Jadhav , 22, India

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

3 solutions

Surya Prakash
Jul 31, 2015

∑ n = 1 25 2 n − 1 ( n ) ( n + 1 ) ( n + 2 ) \sum_{n=1}^{25} \dfrac{2n-1}{(n)(n+1)(n+2)} = ∑ n = 1 25 − 1 2 n + 3 n + 1 + − 5 2 ( n + 2 ) =\sum_{n=1}^{25} \dfrac{-1}{2n} + \dfrac{3}{n+1} + \dfrac{-5}{2(n+2)} = ∑ n = 1 25 − 1 2 n + ∑ n = 1 25 3 n + 1 + ∑ n = 1 25 − 5 2 ( n + 2 ) =\sum_{n=1}^{25} \dfrac{-1}{2n} +\sum_{n=1}^{25} \dfrac{3}{n+1} + \sum_{n=1}^{25} \dfrac{-5}{2(n+2)}

Let S = ∑ n = 1 25 1 n S= \sum_{n=1}^{25} \dfrac{1}{n} .

So,

= − 1 2 ( S ) + 3 ( S + 1 26 − 1 ) − 5 2 ( S + 1 27 + 1 26 − 1 2 − 1 ) = \dfrac{-1}{2} ( S ) + 3 \left( S+ \dfrac{1}{26} - 1 \right) - \dfrac{5}{2} \left( S + \dfrac{1}{27} + \dfrac{1}{26} - \dfrac{1}{2} - 1\right) = 475 702 = \dfrac{475}{702}

So, M = 475 M = 475 and E = 702 E = 702 .

Therefore, M + E = 1177 M+E = \boxed{1177} .

10 Helpful 0 Interesting 0 Brilliant 0 Confused

Moderator note:

It's great that you recognized we could use partial fractions to approach this problem!

Bad calculation three times !! . Up voted your solution. Can you please explain your method after "Let S=...." ........Any way my approach...
f ( n ) = ∑ n = 1 25 2 n − 1 n ( n + 1 ) ( n + 2 ) . \displaystyle f(n)=\sum_{n=1}^{25} \dfrac{2n-1}{n(n+1)(n+2)}. \\ 2 n − 1 n ( n + 1 ) ( n + 2 ) = 5 ( n + 1 ) − 3 ( n + 2 ) n ( n + 1 ) ( n + 2 ) = 5 n ( n + 2 ) − 3 n ( n + 1 ) \dfrac{2n-1}{n(n+1)(n+2)} = \dfrac{5(n+1)-3(n+2)}{n(n+1)(n+2)} = \dfrac 5{n(n+2)} - \dfrac 3 {n(n+1)} \\ = { 5 2 n − 5 2 ( n + 2 ) } − { 3 n − 3 n + 1 } ∴ f ( n ) = − ∑ n = 1 25 1 2 n + ∑ n = 1 25 3 n + 1 − ∑ n = 1 25 5 2 ( n + 2 ) = \left \{ \dfrac 5{2n} - \dfrac 5{2(n+2)} \right \} -\left \{ \dfrac 3 n - \dfrac 3 {n+1} \right \}\\ \displaystyle \therefore~f(n)= - \sum_{n=1}^{25} \dfrac 1{2n} +\sum_{n=1}^{25} \dfrac 3 {n+1} - \sum_{n=1}^{25} \dfrac 5{2(n+2)} \\
= − 1 2 ∗ ∑ n = 1 2 1 n − 1 2 ∗ ∑ n = 3 25 1 n + { 3 ∗ ∑ n = 2 2 1 n + 3 ∗ ∑ n = 3 25 1 n + 3 ∗ ∑ n = 26 26 1 n } − 5 2 ∗ ∑ n = 3 25 1 n − 5 2 ∗ ∑ n = 26 27 1 n = − 1 2 − 1 4 + { 3 2 + 3 26 } − 5 2 ∗ 1 26 − 5 2 ∗ 1 27 = 475 702 m + n = 1177 \displaystyle = - \frac 1 2 *\sum_{n=1}^2 \dfrac 1 n - \color{#3D99F6}{\frac 1 2 * \sum_{n=3}^{25} \dfrac 1 {n}} +~ \left \{ 3*\sum_{n=2}^ 2 \dfrac 1 n + \color{#3D99F6}{3*\sum_{n=3}^{ 25} \dfrac 1 n }+ 3*\sum_{n=26}^{ 26} \dfrac 1 n \right \} ~\\ \displaystyle - \color{#3D99F6}{ \frac 5 2* \sum_{n=3}^{25} \dfrac 1 n} - \frac 5 2*\sum_{n=26}^{27} \dfrac 1 n \\ = - \dfrac 1 2 - \dfrac 1 4 + \left \{ \dfrac 3 2 + \dfrac 3 { 26} \right \} - \dfrac 5 2* \dfrac 1 {26} - \dfrac 5 2* \dfrac 1 {27} =\dfrac {475}{702}\\ m+n=1177

Niranjan Khanderia - 5 years, 4 months ago

Log in to reply

I did the same. But calculation got wrong.

Akshat Joshi - 4 years, 4 months ago

Log in to reply

In the first two attempts I got my calculation wrong , in the last one with utmost care , I did it and got it right :) :)

Ankit Kumar Jain - 4 years, 2 months ago

There is a typo in the second line of your solution.

Ankit Kumar Jain - 4 years, 2 months ago

Log in to reply

Thank you. I have corrected.

Niranjan Khanderia - 4 years, 1 month ago
汶良 林
Aug 8, 2015

8 Helpful 0 Interesting 0 Brilliant 0 Confused

Moderator note:

This is how I done it! You don't have to completely express the fraction as partial fractions, as long as its in the form of a n + 1 − a n a_{n+1} - a_n .

Did the exact same

Aditya Kumar - 5 years ago

2 n − 1 n ( n + 1 ) ( n + 2 ) = A n + B n + 1 + C n + 2 ⇒ \displaystyle \frac{2n - 1}{n(n + 1)(n + 2)} = \frac{A}{n} + \frac{B}{n + 1} + \frac{C}{n + 2} \Rightarrow 2 n − 1 = A ( n + 1 ) ( n + 2 ) + B ( n 2 + 2 n ) + C ( n 2 + n ) = A ( n 2 + 3 n + 2 ) + B ( n 2 + 2 n ) + C ( n 2 + n ) ⇒ 2n - 1 = A(n +1)(n +2) + B(n^2 + 2n) + C(n^2 + n) = A(n^2 + 3n + 2) + B(n^2 + 2n) + C(n^2 + n) \Rightarrow 2 A = − 1 , 3 A + 2 B + C = 2 , A + B + C = 0 ⇒ 2A = - 1, \space 3A + 2B + C = 2, \space A + B + C = 0 \Rightarrow A = − 1 2 , → B = 3 → C = − 5 2 . A = \frac{-1}{2}, \space \rightarrow B = 3 \space \rightarrow C = \frac{-5}{2}. The rest is similar to Surya Prakash's solution.

0 Helpful 0 Interesting 0 Brilliant 0 Confused

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...