Question

How many terms of series 1+3+9+……, sum to 1093

Answer 1

Hint: As we can clearly observe that the series is a GP (Geometric Progression) and its common ratio is equal to 3. We can clearly use the formula for the sum of GP terms for first n terms and from there determine the value of n for which the sum is equal to 1093. The sum of series up to first n terms may be given by formula: \[{{S}_{n}}=\dfrac{{{3}^{n-1}}-1}{3-1}\], \[{{S}_{n}}=1093\](Given)

Complete step-by-step solution:
The problem like this becomes easy when we develop some intuition of Geometric progression. A series may be represented as \[{{a}_{1}}+{{a}_{2}}+{{a}_{3}}+\ldots +{{a}_{n}}\], where \[{{a}_{n}}\] represents \[nth\] term of the series. So by little bit observation, we can see that in this series i.e. \[1+3+\ldots \] if we divide the following term by its preceding term we get a similar number 3, which is known as the common ratio of the series, which certainly means that each term is common ratio times its previous term. So we can easily derive the general formula for the sum of series up to n terms by a very wise mathematical trick.
\[{{S}_{n}}=a+ar+a{{r}^{2}}+\ldots +a{{r}^{n-1}}\ldots eq(1)\]
Multiplying \[{{S}_{n}}\] by common ratio r
\[\Rightarrow {{S}_{n}}r=ar+a{{r}^{2}}+a{{r}^{3}}+\ldots +a{{r}^{n-1}}+a{{r}^{n}}\ldots eq(2)\]
\[\Rightarrow {{S}_{n}}-{{S}_{n}}r=a-a{{r}^{n}}\]
\[\Rightarrow {{S}_{n}}(1-r)=a-a{{r}^{n}}\]
\[\Rightarrow n=7\]
Multiplying numerator and denominator by -1,the dfraction remains unchanged
\[\Rightarrow {{S}_{n}}=\dfrac{a({{r}^{n}}-1)}{r-1}\]
Now we can focus on series given in the question, as we can see \[1+3+9\ldots {{1.3}^{n-1}}\] which means as per concepts developed above our common ratio is 3 and the initial term is 1. So according to question we need to find that value of n up to which the sum of series comes out to be equals to 1093. Simply by solving for the value of this n gives us our answer. Let us solve this question in a stepwise method.
\[\dfrac{a({{r}^{n}}-1)}{r-1}=1093\]
Putting value of a and r,
\[\Rightarrow \dfrac{1({{3}^{n}}-1)}{3-1}=1093\]
\[\Rightarrow {{3}^{n}}-1=1093\times 2\]
\[\Rightarrow {{3}^{n}}-1=2186\]
\[\Rightarrow {{3}^{n}}=2187={{3}^{7}}\]
\[\Rightarrow n=7\]
Hence, the answer is \[n=7\].

Note: Students should not waste their time cramming the formula written in their textbook but they should try to learn this derivation as we have done in the solution section. At last, a student must learn to apply a different approach to answer the same question and that is possible when they learn to think out of the box.