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

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

Explanation

Solution

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
Sn=∑i=1nai{S_n} = \sum\limits_{i = 1}^n {{a_i}}

Complete step-by-step answer:
The terms of the given sequence are
a1=1 a2=5 a3=12 ⋮  {a_1} = 1 \\\ {a_2} = 5 \\\ {a_3} = 12 \\\ \vdots \\\
We see their differences are in arithmetic progression with first term 4 and common difference 3, i.e.
 a2−a1=4 +a3−a2=7 +a4−a3=10 ⋮ ⋮ +an−an−1=3n−2  {\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 \\\
Adding all the equations, we get
an−a1=4+7+10+......+(3n−2) As 4+7+10+......+(3n−2) are in AP where a=4, d=3 and n′=n−1 So by using the formula of sum of an AP we get, an−1=(n−1)(2(4)+(n−1−1)3)2 On simplifying we get, ⇒an−1=(n−1)(4+3n−2)2=(n−1)(3n+2)2 ⇒an=3n2+2n−3n−2+22=3n2−n2  {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} \\\
Therefore sum up to the nth term of the given sequence is
Sn=∑i=1nai=∑i=1n3i2−i2 ⇒Sn=32∑i2−12∑i Now we use the fact that ∑i2=n(n+1)(2n+1)6and ∑i=n(n+1)2, ⇒Sn=32∗n(n+1)(2n+1)6−12∗n(n+1)2 ⇒Sn=n(n+1)(2n+1)4−n(n+1)4 On simplifying further we get, ⇒Sn=n(n+1)4[2n+1−1]=n2(n+1)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 n2(n+1)2\dfrac{{{n^2}(n + 1)}}{2}

Note: Apparently, the terms of the given sequence have no pattern. Now, note that the differences of the terms of the given sequence are in arithmetic progression. Using this, find the closed expression for the nth term of the sequence first. Then calculate the sum up to nth term
Sn=∑i=1nai{S_n} = \sum\limits_{i = 1}^n {{a_i}}