Find the sum of the series 3+7+14+24+37..

Explanation

Clearly here the differences between the successive terms are
`7 - 3, 14 - 7, 24 - 14 .... " i.e.," 4, 7, 10, 13,....`, which are in A.P.
Let `S = 3 + 7 + 14 + 24 + ... + T_(n)`
`S = 3 + 7 + 14 + ... + T_(n -1) + T_(n)`
Subtracting, we get
`0 = 3 + [4 + 7 + 10 + 13 + ... (n - 1)" terms "] - T_(n)`
`:. T_(n) = 3 + S_(n -1)` of an A.P. whose a = 4 and d = 3
`:. T_(n) = 3 + ((n - 1)/(2)) (2.4 +(n - 2)3) = (6 + (n - 1)(3n + 2))/(4) " or " T_(n) = (1)/(2) (3n^(2) - n + 4)`
Now putting `n = 1, 2, 3`..., n and adding
`:. S_(n) = (1)/(2) [3 sum n^(2) - sum n + 4n ] = (1)/(2) [3(n(n + 1) (2n + 1))/(6) - (n(n + 1))/(2) + 4n] = (n)/(2) (n^(2) + n + 4)`
Aliter Method:
Let `T_(n) = an^(2) + bn + c`
Now, `T_(1) = 3 = a + b + c`...(i)
`T_(2) = 7 = 4a + 2b + c`....(ii)
`T_(3) = 14 = 8a + 3 b + c` ...(iii)
Solving (i), (ii) & (iii) we get
`a= (3)/(2), b = - (1)/(2) " &" c = 2`
`:. T_(n) = (1)/(2) (3n^(2) - n + 4)`
`rArr s_(n) = SigmaT_(n) = (1)/(2) [3 Sigma n^(2) - Sigman + 4n] = (1)/(2) [3(n(n + 1) (2n + 1))/(6) - (n(n + 1))/(2) + 4n] = (n)/(2) (n^(2) + n + 4)`