Question

# Find sum up to n terms of the series 1+5+12+22+35....

Hint: Note that, the differences of the terms of the given sequence are in arithmetic progression. Find the closed expression for the nth term of the sequence first. Then calculate the sum up to the nth term
${S_n} = \sum\limits_{i = 1}^n {{a_i}}$

${a_1} = 1 \\ {a_2} = 5 \\ {a_3} = 12 \\ \vdots \\$
${\text{ }}{{{a}}_2} - {a_1} = 4 \\ {}_ + {{{a}}_3} - {{{a}}_2} = 7 \\ {}_ + {{{a}}_4} - {{{a}}_3} = 10 \\ \vdots \\ \vdots \\ {}_ + {a_n} - {{{a}}_{n - 1}} = 3n - 2 \\$
${a_n} - {a_1} = 4 + 7 + 10 + ...... + (3n - 2) \\ As{\text{ }}4 + 7 + 10 + ...... + (3n - 2){\text{ }}are{\text{ }}in{\text{ }}AP \\ where{\text{ }}a = 4,{\text{ }}d = 3{\text{ }}and{\text{ }}n' = n - 1 \\ So{\text{ }}by{\text{ }} using {\text{ }} the {\text{ }} formula {\text{ }}of{\text{ }}sum{\text{ }}of{\text{ }}an{\text{ }}AP{\text{ }}we{\text{ }}get, \\ {a_n} - 1 = \dfrac{{(n - 1)(2(4) + (n - 1 - 1)3)}}{2} \\ On{\text{ }}simplifying{\text{ }}we{\text{ }}get, \\ \Rightarrow {a_n} - 1 = \dfrac{{(n - 1)(4 + 3n - 2)}}{2} = \dfrac{{(n - 1)(3n + 2)}}{2} \\ \Rightarrow {a_n} = \dfrac{{3{n^2} + 2n - 3n - 2 + 2}}{2} = \dfrac{{3{n^2} - n}}{2} \\$
${S_n} = \sum\limits_{i = 1}^n {{a_i}} = \sum\limits_{i = 1}^n {\dfrac{{3{i^2} - i}}{2}} \\ \Rightarrow {S_n} = \dfrac{3}{2}\sum {{i^2}} - \dfrac{1}{2}\sum i \\ Now{\text{ }}we{\text{ }}use{\text{ }}the{\text{ }}fact{\text{ }}that{\text{ }}\sum {{i^2}} = \dfrac{{n(n + 1)(2n + 1)}}{6}and{\text{ }}\sum i = \dfrac{{n(n + 1)}}{2}, \\ \Rightarrow {S_n} = \dfrac{3}{2}*\dfrac{{n(n + 1)(2n + 1)}}{6} - \dfrac{1}{2}*\dfrac{{n(n + 1)}}{2} \\ \Rightarrow {S_n} = \dfrac{{n(n + 1)(2n + 1)}}{4} - \dfrac{{n(n + 1)}}{4} \\ On{\text{ }}simplifying{\text{ }}further{\text{ }}we{\text{ }}get, \\ \Rightarrow {S_n} = \dfrac{{n(n + 1)}}{4}\left[ {2n + 1 - 1} \right] = \dfrac{{{n^2}(n + 1)}}{2} \\$
Hence, sum up to n terms of the series 1+5+12+22+35.... is given by $\dfrac{{{n^2}(n + 1)}}{2}$
${S_n} = \sum\limits_{i = 1}^n {{a_i}}$