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