Answer

Verified

426.6k+ views

**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.

**Complete step-by-step answer:**

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.

Recently Updated Pages

Which of the following is correct regarding the Indian class 10 social science CBSE

Who was the first sultan of delhi to issue regular class 10 social science CBSE

The Nagarjuna Sagar project was constructed on the class 10 social science CBSE

Which one of the following countries is the largest class 10 social science CBSE

What is Biosphere class 10 social science CBSE

Read the following statement and choose the best possible class 10 social science CBSE

Trending doubts

Fill the blanks with the suitable prepositions 1 The class 9 english CBSE

Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE

Which are the Top 10 Largest Countries of the World?

Give 10 examples for herbs , shrubs , climbers , creepers

Write a letter to the principal requesting him to grant class 10 english CBSE

Why is the Earth called a unique planet class 6 social science CBSE

Change the following sentences into negative and interrogative class 10 english CBSE

One cusec is equal to how many liters class 8 maths CBSE