
Find sum up to n terms of the series 1+5+12+22+35....
Answer
522.6k+ 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}} $
Complete step-by-step answer:
The terms of the given sequence are
$
{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.
$
{\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
$
{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
$
{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 $\dfrac{{{n^2}(n + 1)}}{2}$
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
${S_n} = \sum\limits_{i = 1}^n {{a_i}} $
${S_n} = \sum\limits_{i = 1}^n {{a_i}} $
Complete step-by-step answer:
The terms of the given sequence are
$
{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.
$
{\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
$
{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
$
{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 $\dfrac{{{n^2}(n + 1)}}{2}$
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
${S_n} = \sum\limits_{i = 1}^n {{a_i}} $
Recently Updated Pages
Master Class 11 Computer Science: Engaging Questions & Answers for Success

Power set of empty set has exactly subset class 11 maths CBSE

Master Class 10 Computer Science: Engaging Questions & Answers for Success

Master Class 10 General Knowledge: Engaging Questions & Answers for Success

Master Class 9 General Knowledge: Engaging Questions & Answers for Success

Master Class 9 English: Engaging Questions & Answers for Success

Trending doubts
1 ton equals to A 100 kg B 1000 kg C 10 kg D 10000 class 11 physics CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

How much is 23 kg in pounds class 11 chemistry CBSE

1 Quintal is equal to a 110 kg b 10 kg c 100kg d 1000 class 11 physics CBSE

Whales are warmblooded animals which live in cold seas class 11 biology CBSE
