Question

If $\sum\limits_{r=1}^{n}{I\left( r \right)}={{3}^{n}}-1$, then $\sum\limits_{r=1}^{n}{\dfrac{1}{I\left( r \right)}}$ is equal to

Answer 1

Hint: We first try to find the individual terms of the sequence $I\left( 1 \right),I\left( 2 \right),I\left( 3 \right),..........,I\left( n \right)$. We can form the terms in the expression of sums. We get the general form and find the sequence of its reciprocal forms for $\sum\limits_{r=1}^{n}{\dfrac{1}{I\left( r \right)}}$. Then we find the sum of the new sequence.

Complete step by step solution:
First, we try to find the terms of the expression of the sum $\sum\limits_{r=1}^{n}{I\left( r \right)}={{3}^{n}}-1$.
The terms are $I\left( 1 \right),I\left( 2 \right),I\left( 3 \right),..........,I\left( n \right)$.
Changing the value of $n$ in $\sum\limits_{r=1}^{n}{I\left( r \right)}$, we can find the sum of the required number of terms. If we assume ${{S}_{n}}=\sum\limits_{r=1}^{n}{I\left( r \right)}$, we can form the terms in the expression of sums.
Therefore, ${{S}_{1}}=I\left( 1 \right),{{S}_{2}}=\sum\limits_{r=1}^{2}{I\left( r \right)}=I\left( 1 \right)+I\left( 2 \right),{{S}_{3}}=\sum\limits_{r=1}^{3}{I\left( r \right)}=I\left( 1 \right)+I\left( 2 \right)+I\left( 3 \right),......$
We can write ${{S}_{1}}=I\left( 1 \right),I\left( 2 \right)={{S}_{2}}-{{S}_{1}},I\left( 3 \right)={{S}_{3}}-{{S}_{2}},......$
The general form being $I\left( n \right)={{S}_{n}}-{{S}_{n-1}},n\ge 2$.
Now we find the terms where we have
\[\begin{align}
  & I\left( 1 \right)={{S}_{1}}={{3}^{1}}-1=2 \\
 & I\left( 2 \right)={{S}_{2}}-{{S}_{1}}=\left( {{3}^{2}}-1 \right)-\left( {{3}^{1}}-1 \right)=6 \\
 & I\left( 3 \right)={{S}_{3}}-{{S}_{2}}=\left( {{3}^{3}}-1 \right)-\left( {{3}^{2}}-1 \right)=18 \\
 & I\left( 4 \right)={{S}_{4}}-{{S}_{3}}=\left( {{3}^{4}}-1 \right)-\left( {{3}^{3}}-1 \right)=54 \\
\end{align}\]
We can see the terms 2, 6, 18, 54 form a G.P. series. The common ratio is $k=\dfrac{6}{2}=\dfrac{18}{6}=3$.
The general term will be $I\left( r \right)=I\left( 1 \right){{k}^{r-1}}$ which is equal to $I\left( r \right)=2\times {{3}^{r-1}}$.
Therefore, $\dfrac{1}{I\left( r \right)}=\dfrac{1}{2\times {{3}^{r-1}}}$. The first term will be $\dfrac{1}{I\left( 1 \right)}=\dfrac{1}{2}$.
The sequence of $\dfrac{1}{I\left( r \right)}=\dfrac{1}{2\times {{3}^{r-1}}}$ is also a G.P. with its common ratio being $m=\dfrac{1}{3}$.
We need to find the sum of n terms as $\sum\limits_{r=1}^{n}{\dfrac{1}{I\left( r \right)}}$.
The value of $\left| m \right|=\dfrac{1}{3}<1$ for which the sum of the first n terms of an G.P. is ${{s}_{n}}=\dfrac{1}{I\left( 1 \right)}\times \dfrac{1-{{r}^{n}}}{1-r}$.
The summation notation for the series \[\dfrac{1}{2},\dfrac{1}{6},\dfrac{1}{18},...,\dfrac{1}{2\times {{3}^{r-1}}}\] will be ${{s}_{n}}=\dfrac{1}{2}\times \dfrac{1-{{\left( \dfrac{1}{2} \right)}^{n}}}{1-\left( \dfrac{1}{2} \right)}=1-\dfrac{1}{{{2}^{n}}}$.
Therefore, $\sum\limits_{r=1}^{n}{\dfrac{1}{I\left( r \right)}}$ is equal to $1-\dfrac{1}{{{2}^{n}}}$.

Note: The second sequence is a decreasing sequence where the common ratio is a positive number. The common difference will never be calculated according to the difference of greater number from the lesser number. The ratio formula should always be according $r=\dfrac{{{t}_{2}}}{{{t}_{1}}}=\dfrac{{{t}_{3}}}{{{t}_{2}}}=\dfrac{{{t}_{4}}}{{{t}_{3}}}$ in its general form.