Courses for Kids
Free study material
Offline Centres
Store Icon

Solve ${{2.2}^{2}}+{{22.2}^{2}}+{{222.2}^{2}}........$ up to n – term.

Last updated date: 13th Jun 2024
Total views: 403.5k
Views today: 11.03k
403.5k+ views
Hint: To solve this question, firstly we will take the common terms out of the series and we will then have summation of 1+11+111+… up to n in brackets. Then, we will rearrange the terms and will then use the formula to find the summation of n terms of GP. And then on simplification, we will have the solution of the question.

Complete step by step answer:
First let us do some modification and re - arrangement in the series.
Now, in question, the series we are provided is ${{2.2}^{2}}+{{22.2}^{2}}+{{222.2}^{2}}........$
Let summation of the given series be denoted by ${{S}_{n}}$ .
${{S}_{n}}={{2.2}^{2}}+{{22.2}^{2}}+{{222.2}^{2}}........$, where ${{n}^{2}}$ denotes the ${{n}^{th}}$ term of the summation.
Now, let solve this question.
As, we can see that all the terms are multiple of 2, such as ${{2}^{2}}={{2}^{2}}.2$, ${{22}^{2}}={{2.2}^{2}}.11$
So firstly, let us take out ${{2}^{2}}$ common out from al terms of the summation of series.
So, we get
Now, let us multiply and divide the bracket by factor 9, we get
So, on multiplying all the terms by 9 in the bracket, we get
Now, as we can see that we can write 9 = 10 – 1, $99={{10}^{2}}-1$, $999={{10}^{3}}-1$,….
So, on substituting terms similarly in above equation, we get
${{S}_{n}}=2.\dfrac{{{2}^{2}}}{9}[\left( 10-1 \right)+\left( {{10}^{2}}-1 \right)+\left( {{10}^{3}}-1 \right)+........+n]$
Now, re – arranging all the similar terms or same kind of terms together we get
${{S}_{n}}=2.\dfrac{{{2}^{2}}}{9}\left[ \left( 10+{{10}^{2}}+{{10}^{3}}+......+n \right)-(1+1+1+.....+n) \right]$
As we know that if we add a same number n times we will get n times of that number, which means if we add a n times a + a + …. + a, so we get na.
Here , we have a = 1, so $1+1+1+.....+1=n$.
So, we get
${{S}_{n}}=2.\dfrac{{{2}^{2}}}{9}\left[ \left( 10+{{10}^{2}}+{{10}^{3}}+......+n \right)-n \right]$
Also, in $10+{{10}^{2}}+{{10}^{3}}+......+n$, we can se the together forms GP
So, we know that sum of n terms of G.P is equals to ${{S}_{n}}=a\left( \dfrac{{{a}^{n}}-1}{a-1} \right)$, where a is first term of G.P and n is number of terms in GP.
In $10+{{10}^{2}}+{{10}^{3}}+......+n$, a = 10 and n = n
So, summation of $10+{{10}^{2}}+{{10}^{3}}+......+n$ is equals to $10+{{10}^{2}}+{{10}^{3}}+......+n=10\left( \dfrac{{{10}^{n}}-1}{10-1} \right)$
So, we get
${{S}_{n}}=2.\dfrac{{{2}^{2}}}{9}\left[ 10\left( \dfrac{{{10}^{n}}-1}{10-1} \right)-n \right]$
On simplifying, we get
${{S}_{n}}=2.\dfrac{{{2}^{2}}}{9}\left[ 10\left( \dfrac{{{10}^{n}}-1}{9} \right)-n \right]$
Hence, ${{S}_{n}}={{2.2}^{2}}+{{22.2}^{2}}+{{222.2}^{2}}........=2.\dfrac{{{2}^{2}}}{9}\left[ 10\left( \dfrac{{{10}^{n}}-1}{9} \right)-n \right]$

Note: Always be careful when you rearrange the terms, the signs are properly assigned to the terms as if the sign gets changed, the whole answer will get wrong. Also, remember that summation of n terms of GP is equals to ${{S}_{n}}=a\left( \dfrac{{{a}^{n}}-1}{a-1} \right)$, where a is the first term of GP and n is the total count of terms in GP. Try not to make any calculation errors.