A growth of triplets

Algebra Level 4

S = 1 1 ⋅ 2 ⋅ 3 + 1 2 ⋅ 3 ⋅ 4 + 1 3 ⋅ 4 ⋅ 5 + 1 4 ⋅ 5 ⋅ 6 + 1 5 ⋅ 6 ⋅ 7 + ⋯ S=\dfrac { 1 }{ 1\cdot 2\cdot 3 } +\dfrac { 1 }{ 2\cdot 3\cdot 4 } +\dfrac { 1 }{ 3\cdot 4\cdot 5 } +\dfrac { 1 }{ 4\cdot 5\cdot 6 } +\dfrac { 1 }{ 5\cdot 6\cdot 7 } +\cdots

If S S can be expressed in the form 1 A \frac { 1 }{ A } with A A being a positive, composite integer, find A A .


The answer is 4.

View wiki
Jonas Katona by Jonas Katona , 22, USA

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

Akshat Sharda
Dec 14, 2015

S = 1 1 ⋅ 2 ⋅ 3 + 1 2 ⋅ 3 ⋅ 4 + … = ∑ n = 1 ∞ 1 n ( n + 1 ) ( n + 2 ) = 1 2 ∑ n = 1 ∞ ( 1 n − 2 n + 1 + 1 n + 3 ) = 1 2 ( ∑ n = 1 ∞ 1 n − 2 ∑ n = 1 ∞ 1 n + 1 + ∑ n = 1 ∞ 1 n + 2 ) = 1 2 ( ∑ n = 1 ∞ 1 n − 2 ∑ n = 2 ∞ 1 n + ∑ n = 3 ∞ 1 n ) = 1 2 ( 1 1 + 1 2 + ∑ n = 3 ∞ 1 n − 2 ⋅ 1 2 − 2 ∑ n = 3 ∞ 1 n + ∑ n = 3 ∞ 1 n ) = 1 2 ( 1 1 + 1 2 − 2 ⋅ 1 2 + ∑ n = 3 ∞ 1 n − 2 ∑ n = 3 ∞ 1 n + ∑ n = 3 ∞ 1 n ) = 1 2 ( 1 + 1 2 − 1 ) = 1 4 \begin{aligned} S & =\frac{1}{1\cdot 2\cdot 3}+\frac{1}{2\cdot 3\cdot 4}+\ldots \\ & = \displaystyle \sum^{\infty}_{n=1}\frac{1}{n(n+1)(n+2)} \\ & = \frac{1}{2} \displaystyle \sum^{\infty}_{n=1}\left(\frac{1}{n}-\frac{2}{n+1}+\frac{1}{n+3}\right) \\ & = \frac{1}{2} \left(\displaystyle \sum^{\infty}_{n=1}\frac{1}{n}-2 \displaystyle \sum^{\infty}_{n=1}\frac{1}{n+1}+ \displaystyle \sum^{\infty}_{n=1}\frac{1}{n+2}\right) \\ & = \frac{1}{2} \left( \displaystyle \sum^{\infty}_{n=1}\frac{1}{n}-2 \displaystyle \sum^{\infty}_{n=2}\frac{1}{n}+ \displaystyle \sum^{\infty}_{n=3}\frac{1}{n}\right) \\ & = \frac{1}{2} \left(\frac{1}{1}+\frac{1}{2} +\displaystyle \sum^{\infty}_{n=3}\frac{1}{n}-2\cdot \frac{1}{2} -2 \displaystyle \sum^{\infty}_{n=3}\frac{1}{n}+ \displaystyle \sum^{\infty}_{n=3}\frac{1}{n}\right) \\ & = \frac{1}{2} \left(\frac{1}{1}+\frac{1}{2}-2\cdot \frac{1}{2} +\displaystyle \sum^{\infty}_{n=3}\frac{1}{n} -2 \displaystyle \sum^{\infty}_{n=3}\frac{1}{n}+ \displaystyle \sum^{\infty}_{n=3}\frac{1}{n}\right) \\ & = \frac{1}{2}\left(1+\frac{1}{2}-1\right) \\ & = \frac{1}{4} \end{aligned}

13 Helpful 0 Interesting 0 Brilliant 0 Confused

https://brilliant.org/problems/infinite-summation-2/?group=uPEto6utdcpg&ref_id=1091372

i've included a generalization of this problem :)

Hamza A - 5 years, 3 months ago

Why is this under calculus? Theres a quite simple algebraic solution to this

Shreyash Rai - 5 years, 5 months ago

Log in to reply

An administrator changed it to algebra...But idk, this kind of telescoping series is under the calculus wiki page

Jonas Katona - 5 years, 5 months ago
Masbahul Islam
Jan 4, 2016

nth term Un=1/n(n+1)(n+2)

so Sn=C-1/2(n+1)(n+2)

putting n=0, Sn=0 so, C=1/4

So, Sn=1/4 - 1/2(n+1)(n+2)

when n tends to infinity Sn=1/4

A=4

1 Helpful 0 Interesting 0 Brilliant 0 Confused
Zeeshan Ali
Dec 19, 2015

I just followed the following algorithm; 

1
2
3
4
5
 
sum <- 0
for i from 1 to inf
    temp <- 1/(i*(i+1)*(i+2))
    sum += temp
return 1/sum

1 Helpful 0 Interesting 0 Brilliant 0 Confused

I have a better way we can multiply and divide by 2 and simplify the term in the following way. 2/2n(n+1)(n+2)={(n+2)-(n)}/2n(n+1)(n+2) Now by simplifying with the help of Telescopic sum we can cancel each term Final term=n(n+3)/4(n+1)(n+2).. Now use L'hopital 's rule to calculate the limit and approaches infinity

Aman Kumar - 3 years, 1 month ago

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...