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Question: If the sum of n terms of a G.P. denoted by \({{S}_{n}}\) whose first term is a, and common ratio r, ...

If the sum of n terms of a G.P. denoted by Sn{{S}_{n}} whose first term is a, and common ratio r, find sum of S1,S2,S3,........S2n−1.{{S}_{1}},{{S}_{2}},{{S}_{3}},........{{S}_{2n-1}}.

Explanation

Solution

Hint: Given sequence is a geometric progression. So, apply the sum of n terms formula and find the value of all terms in required expression. Then do the sum of all the results obtained by solving each term. Take the term which is common from the bracket and simplify the expression. By doing this you will again get a geometric sequence inside the bracket. By applying the formula again, you get a simplified expression which is our required result. If there is a geometric progression with first term ‘a’ and common ratio “r”. The sum of n terms is given by:
Sn=a(rn−1)r−1{{S}_{n}}=\dfrac{a\left( {{r}^{n}}-1 \right)}{r-1}

Complete step-by-step answer:
Given expression in the question for which we need to find the value:
S1+S3+S5.......+S2n−1{{S}_{1}}+{{S}_{3}}+{{S}_{5}}.......+{{S}_{2n-1}}
By basic knowledge of geometric progression, we can write sum as:
Case-1: Substituting the value of n=1, we get:
S1=a(r−1)r−1=a{{S}_{1}}=\dfrac{a\left( r-1 \right)}{r-1}=a
Case-2: By substituting the value of n as 3, we get:
S3=a(r3−1)r−1{{S}_{3}}=\dfrac{a\left( {{r}^{3}}-1 \right)}{r-1}
Similarly, by substituting n as 2n-1, we get:
S2n−1=a(r2n−1−1)r−1{{S}_{2n-1}}=\dfrac{a\left( {{r}^{2n-1}}-1 \right)}{r-1}
By adding all such cases above, we get the expression as:
S1+S3+S5.......+S2n−1=a(r−1)r−1+a(r3−1)r−1+..........a(rn−1−1)r−1{{S}_{1}}+{{S}_{3}}+{{S}_{5}}.......+{{S}_{2n-1}}=a\dfrac{\left( r-1 \right)}{r-1}+\dfrac{a\left( {{r}^{3}}-1 \right)}{r-1}+..........\dfrac{a\left( {{r}^{n-1}}-1 \right)}{r-1}
By taking the terms “a”, “r-1” common, we get it as:
S1+S3+S5.......+S2n−1=ar−1((r−1)+(r3−1)+.........+(r2n−1−1)){{S}_{1}}+{{S}_{3}}+{{S}_{5}}.......+{{S}_{2n-1}}=\dfrac{a}{r-1}\left( \left( r-1 \right)+\left( {{r}^{3}}-1 \right)+.........+\left( {{r}^{2n-1}}-1 \right) \right)
By grouping “r” terms, -1’s separately, we get the expression as:
S1+S3+...........+S2n−1=ar−1((r+r3+......+r2n−1)−n){{S}_{1}}+{{S}_{3}}+...........+{{S}_{2n-1}}=\dfrac{a}{r-1}\left( \left( r+{{r}^{3}}+......+{{r}^{2n-1}} \right)-n \right)
The equation inside the bracket is geometric progression with first term r and common ratio r2{{r}^{2}}.
By applying sum of n terms to above equation, we get results as follow:

& {{S}_{1}}+{{S}_{3}}+...........+{{S}_{2n-1}}=\dfrac{a}{r-1}\left( r\dfrac{\left( {{\left( {{r}^{2}} \right)}^{n}}-1 \right)}{{{r}^{2}}-1}-n \right) \\\ & {{S}_{1}}+{{S}_{3}}+...........+{{S}_{2n-1}}=\dfrac{a}{r-1}\left( r\dfrac{\left( {{r}^{2n}}-1 \right)}{{{r}^{2}}-1}-n \right) \\\ \end{aligned}$$ Therefore, the above equation is the required result in the question. Note: The idea of applying the sum of n terms of geometric progression ${2}^{nd}$ time is for making the result look simpler with only finite terms for calculations. While applying it, be careful to write n not (n-1).