Courses
Courses for Kids
Free study material
Offline Centres
More
Store

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

Last updated date: 20th Jun 2024
Total views: 413.7k
Views today: 7.13k
Verified
413.7k+ views
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 $\dfrac{{{n^2}(n + 1)}}{2}$
${S_n} = \sum\limits_{i = 1}^n {{a_i}}$