Question

What is $\dfrac{1}{1} + \dfrac{1}{{1 + 2}} + \dfrac{1}{{1 + 2 + 3}} + ....... + \dfrac{1}{{1 + 2 + 3 + ... + 2015}}$?

Answer 1

To find the value of $\dfrac{1}{1} + \dfrac{1}{{1 + 2}} + \dfrac{1}{{1 + 2 + 3}} + ....... + \dfrac{1}{{1 + 2 + 3 + ... + 2015}}$ we will first solve for general term ‘$n$’ by finding the general ${n^{th}}$ term and then using ${s_n} = \sum\limits_{k = 1}^{k = n} {{t_k}}$ we will find the sum in terms of $n$. At last, we will put $n = 2015$ to find the required result.

Complete step-by-step solution:
We have to find the value of $\dfrac{1}{1} + \dfrac{1}{{1 + 2}} + \dfrac{1}{{1 + 2 + 3}} + ....... + \dfrac{1}{{1 + 2 + 3 + ... + 2015}}$.
Here, we will use the concept of summation. The summation is a process of adding up a sequence of given numbers, the result is their sum or total. It is usually required when the large numbers of data are given and it instructs to total up all values in a given sequence.
Let ${s_n} = \dfrac{1}{1} + \dfrac{1}{{1 + 2}} + \dfrac{1}{{1 + 2 + 3}} + ....... + \dfrac{1}{{1 + 2 + 3 + ... + 2015}}$.
Also, let any general ${n^{th}}$ term as ${t_n}$ which is given by
$\Rightarrow {t_n} = \dfrac{1}{{1 + 2 + 3 + ... + n}}$
We can write ${t_n}$ as
$\Rightarrow {t_n} = \dfrac{1}{{\sum\limits_{k = 1}^{k = n} k }}$
As we know, $1 + 2 + 3 + ... + n = \dfrac{{n\left( {n + 1} \right)}}{2}$, using this we can write
$\Rightarrow {t_n} = \dfrac{1}{{\dfrac{{n\left( {n + 1} \right)}}{2}}}$
On rewriting we get,
$\Rightarrow {t_n} = 2\left[ {\dfrac{{\left( {n + 1} \right) - n}}{{n\left( {n + 1} \right)}}} \right]$
On simplification we get,
$\Rightarrow {t_n} = 2\left[ {\dfrac{{\left( {n + 1} \right)}}{{n\left( {n + 1} \right)}} - \dfrac{n}{{n\left( {n + 1} \right)}}} \right]$
On further simplification we get
$\Rightarrow {t_n} = 2\left[ {\dfrac{1}{n} - \dfrac{1}{{\left( {n + 1} \right)}}} \right]$
As ${S_n} = \sum\limits_{k = 1}^{k = n} {{t_k}}$
Therefore, on putting the values of $k$ from $1$ to $n$ in right side of the above equation, we get
$\Rightarrow {S_n} = {t_1} + {t_2} + {t_3} + ... + {t_n}$
On putting values of $n$ from $1$ to $n$ in ${t_n}$, we get
$\Rightarrow {S_n} = 2\left[ {\left( {\dfrac{1}{1} - \dfrac{1}{2}} \right) + \left( {\dfrac{1}{2} - \dfrac{1}{3}} \right) + \left( {\dfrac{1}{3} - \dfrac{1}{4}} \right) + ... + \left( {\dfrac{1}{{n - 1}} - \dfrac{1}{n}} \right) + \left( {\dfrac{1}{n} - \dfrac{1}{{n + 1}}} \right)} \right]$
On simplification we get,
$\Rightarrow {S_n} = 2\left( {1 - \dfrac{1}{{n + 1}}} \right)$
On taking the LCM, we get
$\Rightarrow {S_n} = 2\left( {\dfrac{{n + 1 - 1}}{{n + 1}}} \right)$
On simplification we get
$\Rightarrow {S_n} = \dfrac{{2n}}{{n + 1}}$
We require a sum of terms up to $n = 2015$ i.e., ${S_{2015}}$ is required.
Therefore putting $n = 2015$, we get
$\Rightarrow {S_{2015}} = \dfrac{{\left( {2 \times 2015} \right)}}{{\left( {2015 + 1} \right)}}$
On simplification we get
$\Rightarrow {S_{2015}} = \dfrac{{2015}}{{1008}}$
Therefore, $\dfrac{1}{1} + \dfrac{1}{{1 + 2}} + \dfrac{1}{{1 + 2 + 3}} + ....... + \dfrac{1}{{1 + 2 + 3 + ... + 2015}}$ is $\dfrac{{2015}}{{1008}}$

Note: Here, we have arranged the terms in such a fashion that it cancels each other and we get simplified results. Also, we have used the concept of summation as it enables us to write short forms for the addition of very large numbers for a given data in a sequence which reduces the complexity of the problem .