Factorials and sums and algebra and .............

Algebra Level 3

1 1 ⋅ 2 + 2 1 ⋅ 2 ⋅ 3 + 3 1 ⋅ 2 ⋅ 3 ⋅ 4 + 4 1 ⋅ 2 ⋅ 3 ⋅ 4 ⋅ 5 + ⋯ \frac 1{1\cdot 2} + \frac 2{1\cdot 2\cdot 3} + \frac 3{1\cdot 2\cdot 3 \cdot 4} + \frac 4{1\cdot 2\cdot 3 \cdot 4 \cdot 5} + \cdots

Find the sum above up to 2018 terms.

1 + 1 2019 ! 1+ \frac 1{2019!} 1 − 1 2018 ! 1- \frac 1{2018!} 1 2018 ! \frac 1{2018!} 1 − 1 2019 ! 1- \frac 1{2019!}

Arghyadeep Chatterjee by Arghyadeep Chatterjee , 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.

2 solutions

Similar solution with @Arghaydeep Chatterjee 's.

S n = 1 1 ⋅ 2 + 2 1 ⋅ 2 ⋅ 3 + 3 1 ⋅ 2 ⋅ 3 ⋅ 4 + ⋯ + n ( n + 1 ) ! = ∑ k = 1 n k ( k + 1 ) ! = ∑ k = 1 n k + 1 − 1 ( k + 1 ) ! = ∑ k = 1 n ( 1 k ! − 1 ( k + 1 ) ! ) = ∑ k = 1 n 1 k ! − ∑ k = 2 n + 1 1 k ! = 1 1 ! − 1 ( n + 1 ) ! \begin{aligned} S_n & = \frac 1{1\cdot 2} + \frac 2{1\cdot 2\cdot 3} + \frac 3{1\cdot 2\cdot 3 \cdot 4} + \cdots + \frac n{(n+1)!} \\ & = \sum_{k=1}^n \frac k{(k+1)!} \\ & = \sum_{k=1}^n \frac {k+1-1}{(k+1)!} \\ & = \sum_{\color{#3D99F6}k=1}^{\color{#3D99F6}n} \left(\frac 1{k!} - \frac 1{(k+1)!}\right) \\ & = \sum_{\color{#3D99F6}k=1}^{\color{#3D99F6}n} \frac 1{\color{#3D99F6}k!} - \sum_{\color{#D61F06}k=2}^{\color{#D61F06}n+1} \frac 1{\color{#D61F06}k!} \\ & = \frac 1{1!} - \frac 1{(n+1)!} \end{aligned}

Putting n = 2018 n=2018 , we have S 2018 = 1 − 1 2019 ! S_{2018} = \boxed{1-\dfrac 1{2019!}} .

4 Helpful 0 Interesting 0 Brilliant 0 Confused

Your solution is more general as it gives you the result for any n belonging to Z. To everyone who stumped at this prob I would advice you to follow @Chew-Seong Cheong's solution rather than mine in order to understand it better.

Arghyadeep Chatterjee - 3 years, 2 months ago

Log in to reply

Thanks for you comment, I was trying to encourage you to use LaTex for writing problem and solution. I have edited this problem for you. So more member will be attracted to solve it.

Chew-Seong Cheong - 3 years, 2 months ago

Log in to reply

I actually do not know how to use LaTex . So I use daum equation writer and then just post the image. Can you perhaps link me a video which explains it.

Arghyadeep Chatterjee - 3 years, 2 months ago

Log in to reply

@Arghyadeep Chatterjee – Instead of pasting the image, you can just learn the codes. You can see LaTex codes by placing your mouse cursor on the formulas in Brilliant.org website. Or click the " ⋯ \cdots More" pull-down menu and select "Toggle LaTex".

Chew-Seong Cheong - 3 years, 2 months ago

3 Helpful 0 Interesting 0 Brilliant 0 Confused

It is a telescoping sum

Vijay Simha - 1 year, 10 months ago

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...