SOLUTION: Find the sum of
(1+2+3....+x) + (2+3+4....+x) + (3+4+5....+x) + ...................... till n terms where n<x
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-> SOLUTION: Find the sum of
(1+2+3....+x) + (2+3+4....+x) + (3+4+5....+x) + ...................... till n terms where n<x
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Question 892087: Find the sum of
(1+2+3....+x) + (2+3+4....+x) + (3+4+5....+x) + ...................... till n terms where n Answer by robertb(5830) (Show Source):
You can put this solution on YOUR website! Arrange the sums in this manner:
1 + 2 + 3 + 4 + ... + n + (n+1) + ... + x
2 + 3 + 4 + ... + n + (n+1) + ... + x
3 + 4 + ... + n + (n+1) + ... + x
4 + ... + n + (n+1) + ... + x
...
(n-1)+n + (n+1) + ... + x
n + (n+1) + ... + x
The triangular portion of the sums can be expressed as
1 + 2*2 + 3*3 + 4*4 +... + (n-1)*(n-1) + n*n, or +...+
Now the sum (n+1)+...+x is added n times. Hence
the sum , by using the sum of an arithmetic sequence.
Therefore, (1+2+3....+x) + (2+3+4....+x) + (3+4+5....+x) + .... till n terms where n < x, is
