Question

The sum of \[n\] terms of the series is \[\dfrac{4}{3} + \dfrac{{10}}{9} + \dfrac{{28}}{{27}} + ...\] is

Answer 1

Hint: In this question, the sum of n terms of the series is given as \[\dfrac{4}{3} + \dfrac{{10}}{9} + \dfrac{{28}}{{27}} + ...\]We need to convert the given terms into particular series where we can apply some formula to get the desired result.

Formula used:
We have been using the following formulas:
\[{S_n} = \dfrac{{a\left( {1 - {r^n}} \right)}}{{1 - r}}\]

Complete step-by-step solution:
 Given that \[\dfrac{4}{3} + \dfrac{{10}}{9} + \dfrac{{28}}{{27}} + ...\]
Now we break the given series as
\[\dfrac{4}{3} + \dfrac{{10}}{9} + \dfrac{{28}}{{27}} + ... = \left( {1 + \dfrac{1}{3}} \right) + \left( {1 + \dfrac{1}{9}} \right) + \left( {1 + \dfrac{1}{{27}}} \right) + \ldots \]
Now we separate inside brackets, we obtain
\[\dfrac{4}{3} + \dfrac{{10}}{9} + \dfrac{{28}}{{27}} + ... = \left( {1 + 1 + 1 + ...n} \right) + \,\left( {\dfrac{1}{3} + \dfrac{1}{9} + \dfrac{1}{{27}} + ...n\,terms} \right)\]
Now we clearly see that the second term in the bracket is G.P
So, we apply the sum of the series formula in G.P which is \[{S_n} = \dfrac{{a\left( {1 - {r^n}} \right)}}{{1 - r}}\] , and by simplifying we get
\[
  \dfrac{4}{3} + \dfrac{{10}}{9} + \dfrac{{28}}{{27}} + ... = n + \dfrac{{\dfrac{1}{3}\left( {1 - \dfrac{1}{{{3^n}}}} \right)}}{{1 - \dfrac{1}{3}}} \\
   = n + \dfrac{1}{2}\left( {1 - \dfrac{1}{{{3^n}}}} \right) \\
   = n + \dfrac{{{3^n} - 1}}{{2 \times {3^n}}} \\
   = \dfrac{{2 \cdot n\,{3^n} + {3^n} - 1}}{{2 \times {3^n}}} \\
 \]
Further simplifying, we get
\[\dfrac{4}{3} + \dfrac{{10}}{9} + \dfrac{{28}}{{27}} + ... = \dfrac{{\left( {2n + 1} \right) \times {3^n} - 1}}{{2\left( {{3^n}} \right)}}\]
Therefore, the sum of \[n\] terms of the series is \[\dfrac{4}{3} + \dfrac{{10}}{9} + \dfrac{{28}}{{27}} + ...\] are \[\dfrac{{\left( {2n + 1} \right) \times {3^n} - 1}}{{2\left( {{3^n}} \right)}}\]

Additional information: A geometric progression is a number sequence in which the ratio of any two consecutive terms is always the same. Simply put, the next number in the series is calculated by multiplying a fixed number by the previous number in the series. The common ratio is the name given to this fixed number.
For example, 2,4,8,16 is a Geometric progression as it has the same difference.

Note: Students should follow the step-by-step procedure just as shown above in the solution and also remember the formula of the sum of series in which the sum of the first “n” terms in a geometric progression with the first term as “a” and common ratio as “r” is given by \[{S_n} = \dfrac{{a\left( {1 - {r^n}} \right)}}{{1 - r}}\]