find the sum of the series upto n terms :
1+9+24+46+....+n
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This has a common second difference=7, so it's a quadratic sequence.
nth term: 3.5n² -2.5n
The first term is 3.5 times the sum of squares, i.e.
(3.5 / 6) n(n+1)(2n+1)
The second is: -2.5 times the sum of integers, i.e.:
(-2.5/2)n(n+1)
So adding we have:
(3.5 / 6)n(n+1)(2n+1) + (-2.5/2)n(n+1)
S(n) = (1/6)(n)(7n²+3n-4)
This checks out S(1)=1, S(2)=10, S(3)=34
If you want, to prove the claim, you could use mathematical induction.
nth term: 3.5n² -2.5n
The first term is 3.5 times the sum of squares, i.e.
(3.5 / 6) n(n+1)(2n+1)
The second is: -2.5 times the sum of integers, i.e.:
(-2.5/2)n(n+1)
So adding we have:
(3.5 / 6)n(n+1)(2n+1) + (-2.5/2)n(n+1)
S(n) = (1/6)(n)(7n²+3n-4)
This checks out S(1)=1, S(2)=10, S(3)=34
If you want, to prove the claim, you could use mathematical induction.
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