An interesting sum

Algebra Level 3

Evaluate 1 + ( 1 + 2 ) + ( 1 + 2 + 3 ) + ⋯ + ( 1 + 2 + 3 + ⋯ + 20 ) . 1+(1+2)+(1+2+3)+\cdots +(1+2+3+\cdots +20).


The answer is 1540.

Sandeep Mehra by Sandeep Mehra , 23, India

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4 solutions

Sunil Pradhan
Oct 23, 2014

series of triangular numbers &

sum of triangular numbers = n(n+1)(n+2)/6

= 20 × 21 × 22/6 = 1540

4 Helpful 0 Interesting 0 Brilliant 0 Confused

@Sunil pradhan i did by another method...... sir, thank you for sharing this formula!

Nihar Mahajan - 6 years, 7 months ago

What are triangular numbers sir?

Chaitnya Shrivastava - 5 years, 9 months ago

A triangular number or triangle number counts the objects that can form an equilateral triangle.

for details https://en.wikipedia.org/wiki/Triangular_number

Sunil Pradhan - 5 years, 9 months ago

The problem becomes easy once you express the required sum as a double sum as follows:

∑ i = 1 20 ( ∑ j = 1 i i ) \large \sum_{i=1}^{20}\left(\sum_{j=1}^i i\right)

Now, you can compute the inner sum and then proceed to compute the outer sum.

Prasun Biswas - 6 years, 4 months ago
Vighnesh Raut
Nov 30, 2014

There are 20 sets...

For an n t h { n }^{ th } set , sum of the numbers in the set is n ( n + 1 ) 2 \frac { n(n+1) }{ 2 }

Also, n varies from 1 to 20 .

So, we have to find ∑ n = 1 20 n ( n + 1 ) 2 \sum _{ n=1 }^{ 20 }{ \frac { n(n+1) }{ 2 } }

= 1 2 ∑ n = 1 20 n 2 + n = 1 2 ( n ( n + 1 ) ( 2 n + 1 ) 6 ) + 1 2 ( n ( n + 1 ) 2 ) w h e r e n = 20 = n ( n + 1 ) ( n + 2 ) 6 w h e r e n = 20 \frac { 1 }{ 2 } \sum _{ n=1 }^{ 20 }{ { n }^{ 2 } } +n\\ =\frac { 1 }{ 2 } \left( \frac { n(n+1)(2n+1) }{ 6 } \right) \quad +\quad \frac { 1 }{ 2 } \left( \frac { n(n+1) }{ 2 } \right) \quad where\quad n=20\\ =\frac { n(n+1)(n+2) }{ 6 } \quad where\quad n=20

So, our answer is 1540

3 Helpful 0 Interesting 0 Brilliant 0 Confused
Ramiel To-ong
Jun 20, 2015

applying the summation techniques: limits from 1 - 20 using n(n+1)/2

0 Helpful 0 Interesting 0 Brilliant 0 Confused
Bala Murugan
Nov 22, 2014

1)find the number of occurences of each number and then multiply by it and then finally add OR 2)do by n(n-1)/2 for each section and then finally add

0 Helpful 0 Interesting 0 Brilliant 0 Confused

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