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Algebra
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@sathyaram
you didn't read the question correctly.
he asked for the sum of the nth group and not the sum of the first n terms..
i am posting the solution.
see a triangle number is of the form n(n+1)/2.
it is the sum of the first n natural numbers. like 1 = 1.
3 = 1 + 2
6 = 3 + 2 + 1
10 = 4 + 3 + 2 + 1 and so on.
we can see that in the question the last element of the nth group is the nth triangle number.
the sum to n groups is then by the formula of AP -
[n(n+1)/4][1 + n(n+1)/2].
for example put n = 4.
we get 5(11) = 55. this is the sum to the 4th group.
now what we need is the sum of nth group.
sum of the nth group = sum to n groups - sum to (n-1) groups.
in the formula above replace n by (n-1) and then subtract.
you will get the answer.....
Just look a group then you will see the following properties
1.The nth group will have n element
2. The last number is given by (n+1)C2
hence the sum of the nth group will be
((n+1)C2 +1)((n+1)C2 )/2-((n+1)C2 -n)((n+1)C2 -n+1)/2
Hope you understand
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use the oncept of triangle numbers.
the last element of each group is a triangle number.
namely = 1,3,6,10.
the last element of the nth group in the nth triangle number.
the nth triangle number = n(n+1)/2.