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Solution
1 + (1 + 3) + (1 + 3 + 5) + ... n terms

Solution:
Your problem `->` 1 + (1 + 3) + (1 + 3 + 5) + ... n terms


`S_n = sum [ f(n) ]`

`= sum [ sum (2n - 1) ]`

`= sum [ 2 sum n - sum 1 ]`

`= sum [ 2 * (n (n + 1))/2 - n ]`

`= sum [ n^2 + n - n ]`

`= sum n^2`

`= (n (n + 1) (2n + 1))/6`






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