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# Find the sum of to n terms of the sequence $\left( {x + \dfrac{1}{{{x^2}}}} \right),\,\,{\left( {{x^2} + \,\,\dfrac{1}{{{x^2}}}} \right)^2},\,{\left( {{x^2} + \,\,\dfrac{1}{{{x^3}}}} \right)^2},.....$.

Last updated date: 03rd Aug 2024
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Hint: An itemized collection of terms in which repetition of any sort of series are allowed is known as sequence.
Sum of n terms of such sequence means adding all the ‘n’ terms.
It could be done with various algebraic expressions
Here, ${(a + b)^2} = {a^2} + {b^2} + 2ab$
Further the concept of A.P & G.P series are used to solve such algebraic expressions
An arithmetic progression is a sequence of numbers such that the difference of any two successive members is a constant. If the initial term of A.P is a, and common difference is ‘d’, then nth terms of sequence is
${a^n} = a + (n - 1)d,$ where n $=$ no. of terms.
Sum of A.P is given by ${S_n} = \dfrac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right],$ where n $=$ no. of terms, a $=$ first term, d$=$common difference.
Geometric sequence has a common ratio.
The formula for the nth term is:
${a_n} = a{r^{n - 1}}$
Where, ${a_n} =$ nth term of the sequence
a$=$first term of the sequence
r$=$common ratio.
Sum of n terms of a G.P is denoted by ${S_n}$i.e.
${S_n} = \dfrac{{a\left( {{r^n} - 1} \right)}}{{(r - 1)}}$
Where, a $=$first term of the sequence.
r$\, =$common ratio.
n $=$ no. of terms in the sequence.

Let ${S_n}$ denote the sum to n terms of the given sequence
Then,
${S_n} = {\left( {x + \dfrac{1}{x}} \right)^2} + {\left( {{x^2} + \dfrac{1}{{{x^2}}}} \right)^2} + {\left( {{x^3} + \dfrac{1}{{{x^2}}}} \right)^2} + ....$
First, we need to find the general term of the sequence i.e. ${\left( {{x^n} + \dfrac{1}{{{x^n}}}} \right)^2}$
Now, ${S_n} = {\left( {x + \dfrac{1}{x}} \right)^2} + {\left( {{x^2} + \dfrac{1}{{{x^2}}}} \right)^2} + {\left( {{x^3} + \dfrac{1}{{{x^2}}}} \right)^2} + ....{\left( {{x^n} + \dfrac{1}{{{x^n}}}} \right)^2}$_____ (1).
Using identity ${\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab$ throughout the sequence (1), we have:
${S_n} = \left[ {{x^2} + \dfrac{1}{{{x^2}}} + 2x \times x} \right] + \left[ {{x^4} + \dfrac{1}{{{x^4}}} + 2{x^2} \times \dfrac{1}{{{x^2}}}} \right] + \left[ {{x^6} + \dfrac{1}{{{x^6}}} + 2{x^6} \times \dfrac{1}{{{x^6}}}} \right] + ......\left[ {{x^{2n}} + \dfrac{1}{{{x^{2n}}}} + 2{x^n} \times \dfrac{1}{{{x^n}}}} \right]$
Simplify the series ${x^2}$ is canceled out.
${S_n} = \left[ {{x^2} + \dfrac{1}{{{x^2}}} + 2} \right] + \left[ {{x^4} + \dfrac{1}{{{x^4}}} + 2} \right] + \left[ {{x^6} + \dfrac{1}{{{x^6}}} + 2} \right] + ......\left[ {{x^{2n}} + \dfrac{1}{{{x^{2n}}}} + 2} \right]$
Combining term, we get:
${S_n} = \left[ {{x^2} + {x^4} + {x^6} + ..... + {x^{2n}}} \right] + \left[ {\dfrac{1}{{{x^2}}} + \dfrac{1}{{{x^4}}} + \dfrac{1}{{x{}^6}} + ......\dfrac{1}{{{x^{2n}}}}} \right] + \left[ {2 + 2 + .........2n} \right]$. ______ (2).
We see that $\left[ {{x^2} + {x^4} + {x^6} + ..... + {x^{2n}}} \right]$ is a G.P with common ratio ${x^2}$and a$= {x^2}$
Sum of finite. G.P $= \dfrac{{a\left( {{r^n} - 1} \right)}}{{r - 1}}$$= \dfrac{{{x^2}\left( {{{\left( {{x^2}} \right)}^n} - 1} \right)}}{{{x^2} - 1}}$
$\dfrac{{{x^2}\left[ {{x^{2n}} - 1} \right]}}{{{x^2} - 1}}$ _______(3)
Now,
$\dfrac{1}{{{x^2}}} + \dfrac{1}{{{x^4}}} + ........ + \dfrac{1}{{{x^{2n}}}}$ is a G.P with a $= \dfrac{1}{{{x^2}}}$ and r $= \dfrac{1}{{{x^2}}}$
Sum of finite G.P $= \dfrac{{a\left( {{r^n} - 1} \right)}}{{r - 1}} = \dfrac{{\dfrac{1}{{{x^2}}}\left( {{{\left( {\dfrac{1}{{{x^2}}}} \right)}^n} - 1} \right)}}{{\left( {\dfrac{1}{{{x^2}}}} \right) - 1}}$ ____________(4).
Substituting equation (3) & (4) in (2), we have:
${S_n} = {x^2}\left[ {\dfrac{{{x^{2n}} - 1}}{{{x^2} - 1}}} \right] + \dfrac{1}{{{x^2}}}\left[ {\dfrac{{{{\left( {\dfrac{1}{{{x^2}}}} \right)}^n} - 1}}{{\left( {\dfrac{1}{{{x^2}}}} \right) - 1}}} \right] + \left( {2 + .... + 2} \right)$n times
n times $2$ will be $2n$
${S_n} = {x^2}\left[ {\dfrac{{{x^{2n}} - 1}}{{{x^2} - 1}}} \right] + \dfrac{1}{{{x^2}}}\left[ {\dfrac{{{{\left( {\dfrac{1}{{{x^2}}}} \right)}^n} - 1}}{{\left( {\dfrac{1}{{{x^2}}}} \right) - 1}}} \right] + 2n$
Simplify
${S_n} = {x^2}\left[ {\dfrac{{{x^{2n}} - 1}}{{{x^2} - 1}}} \right] + \dfrac{1}{{{x^2}}}\left[ {\dfrac{{\dfrac{{1 - {x^{2n}}}}{{{x^{2n}}}}}}{{\dfrac{{1 - {x^2}}}{{{x^2}}}}}} \right] + 2n$
Common denominator is cancelled out.
${S_n} = {x^2}\left[ {\dfrac{{{x^{2n}} - 1}}{{{x^2} - 1}}} \right] + \dfrac{1}{{{x^{2n}}}}\left[ {\dfrac{{1 - {x^{2n}}}}{{1 - {x^2}}}} \right] + 2n$
${S_n} = {x^2}\left[ {\dfrac{{{x^{2n}} - 1}}{{{x^2} - 1}}} \right] + \dfrac{1}{{{x^{2n}}}}\left[ {\dfrac{{ - \left( {{x^{2n}} - 1} \right)}}{{ - \left( {{x^2} - 1} \right)}}} \right] + 2n$
Taking $\left[ {\dfrac{{{x^{2n}} - 1}}{{{x^2} - 1}}} \right]$ common from first $2$terms
${S_n} = \left[ {\dfrac{{{x^{2n}} - 1}}{{{x^2} - 1}}} \right] + \left[ {{x^2} + \dfrac{1}{{{x^{2n}}}}} \right] + 2n$
Hence, it is the required sum.

Note: In such a question all we need is to recognize the sequence whether it is a G.P or an A.P.
Also, as it clearly stated we need to find the sum of n terms i.e. we need to find the nth terms first. Arithmetic progression is of the form:- a-d, a, a+d, a+2d,.....
where a is the first term and
d is the common difference.