Question

Find the value of ${{T}_{50}}$ if ${{T}_{n}}$ denotes the ${{n}^{th}}$ term of the following series:
$2+3+6+11+..............$

Answer 1

Hint: First of all find the difference of the consecutive terms. Then the difference of the consecutive terms are 1, 3, 5,…… Now, as you can see, the difference of the terms given in the question is in A.P. Now, if we add 2 to the series of differences then we get the original series with first term missing. So, the sum of n terms given in the question is equal to addition of 2 with sum of n – 1 terms of the difference series. To find the ${{50}^{th}}$ term of the series, first of all find the sum of 49 terms of the difference series and then add 2 to it.

Complete step-by-step answer:
We have given the following series of which we have to find the summation:
$2+3+6+11+..............$
Taking difference of the consecutive terms we get,
$\begin{align}
  & 3-2=1 \\
 & 6-3=3 \\
 & 11-6=5 \\
\end{align}$
So, the new series that we have obtained from the difference is:
1, 3, 5,…….
Now, to get back the original series we have to add 2 to each term along with the terms preceding to that particular term.
First term is 2.
Second term is calculated by adding 2 with the first term of the difference series (i.e. 1).
$2+1=3$
Hence, the second term is 3.
Now, to calculate the third term of the original series adding 2 with the first two terms of the difference series (i.e. 1 & 3).
$2+1+3=6$
Hence, the third term is 6.
The calculation of the fourth term is calculated by adding 2 with the first three terms of the difference series (i.e. 1, 3, 5).
$2+1+3+5=11$
Hence, the fourth term is 11.
So, from the calculation the original series that we are getting is:
2, 3, 6, 11………
As you can see that the calculated original series is matched with the given one.
So, to find the ${{50}^{th}}$ term we have to add 2 with a sum of 49 terms of the difference series.
Difference series is:
1, 3, 5, 7…………
Sum of 49 terms of the above series is calculated using the formula for summation of a series which is in arithmetic progression.
${{S}_{n}}=\dfrac{n}{2}\left( 2a+\left( n-1 \right)d \right)$
In the above formula, “a” is the first term, “d” is the common difference and n is the number of terms.
In the difference series, the first term (a) is 1 and the common difference is calculated by subtracting any term from its succeeding term. Let us take a term 3 from the difference series and its succeeding term as 5 so subtracting 3 from 5 we get 2. Hence, the common difference (d) is 2.
Number of terms are 49 so the value of n is 49 so substituting the value of a, d and n in the summation formula we get,
$\begin{align}
  & {{S}_{n}}=\dfrac{n}{2}\left( 2a+\left( n-1 \right)d \right) \\
 & \Rightarrow {{S}_{49}}=\dfrac{49}{2}\left( 2\left( 1 \right)+\left( 49-1 \right)2 \right) \\
\end{align}$
Taking 2 as common from the above equation we get,
${{S}_{49}}=\dfrac{49}{2}\left( 2 \right)\left( \left( 1 \right)+\left( 49-1 \right) \right)$
2 will be cancelled out from numerator and denominator we get,
$\begin{align}
  & {{S}_{49}}=\left( 49 \right)\left( \left( 1 \right)+\left( 49-1 \right) \right) \\
 & \Rightarrow {{S}_{49}}={{\left( 49 \right)}^{2}}=2401 \\
\end{align}$
The above calculation gives the sum of 49 terms as 2401. Now, adding 2 into 2401 we get,
$\begin{align}
  & 2401+2 \\
 & =2403 \\
\end{align}$
Hence, the ${{50}^{th}}$ term of the given series in the above problem is 2403.

Note: The place where you go wrong in this question is in writing the original series from the difference series. An instance from the above solution is as follows:
We have to find the ${{50}^{th}}$ term so you might have thought that I can find the ${{49}^{th}}$ term of the difference series and add 2 to it.
The difference series shown above is as follows:
1, 3, 5…..
The above series has first term as 1 and common difference as 2 so the ${{n}^{th}}$ term is:
${{T}_{n}}=a+\left( n-1 \right)d$
Substituting “a” as 1, “d” as 2 in the above equation we get,
${{T}_{n}}=1+\left( n-1 \right)2$
Now, substituting n as 49 in the above formula we get,
$\begin{align}
  & {{T}_{49}}=1+\left( 49-1 \right)2 \\
 & \Rightarrow {{T}_{49}}=1+\left( 48 \right)2 \\
 & \Rightarrow {{T}_{49}}=97 \\
\end{align}$
Adding 2 into 97 will give the ${{50}^{th}}$ term as 99.
As you can see, this is not the ${{50}^{th}}$ term that we are getting above so this way of finding the ${{50}^{th}}$ term is wrong.