Question

If the \[{{n}^{th}}\] term of the series is \[{{T}_{n}}=2n-1\], then sum of n terms \[{{S}_{n}}\] will be: -
(a) \[{{n}^{2}}\]
(b) \[\dfrac{n\left( n+1 \right)}{2}\]
(c) \[\dfrac{n\left( n+1 \right)\left( n+2 \right)}{6}\]
(d) \[n+2\]

Answer 1

Hint: Assume the first term of the given series as ‘a’ and common difference as ‘d’. Find the value of ‘d’ by using the relation: - \[d={{T}_{n}}-{{T}_{n-1}}\], where \[{{T}_{n-1}}\] is the \[{{\left( n-1 \right)}^{th}}\] term of the series which will be obtained by substituting (n – 1) in place of ‘n’ in \[{{T}_{n}}=2n-1\]. Now, apply the formula for sum of ‘n’ term of the A.P. given as: - \[{{S}_{n}}=\dfrac{n}{2}\left[ 2a+\left( n-1 \right)d \right]\] to get the correct answer. Here, \[{{S}_{n}}\] is the sum of ‘n’ terms.

Complete step-by-step solution
Here, we have been provided with the expression for \[{{n}^{th}}\] term of a series, \[{{T}_{n}}=2n-1\] and we have to determine the sum of n terms \[{{S}_{n}}\]. First let us find the type of series the given expression represents. So, we have,
\[\because {{T}_{n}}=2n-1\]
\[\Rightarrow {{T}_{1}}=\] First term = \[2\times 1-1=1\]
\[\Rightarrow {{T}_{2}}=\] second term = \[2\times 2-1=3\]
\[\Rightarrow {{T}_{3}}=\] third term = \[2\times 3-1=5\]
Therefore, the upcoming terms will be 7, 9, 11, 13, ……, and so on. So, on observing the above pattern of consecutive terms, we can say that the series is an arithmetic progression (A.P) whose first term is 1. Now, the common difference of the given A.P. will be the difference between any two consecutive terms. So, we have,
\[\Rightarrow \] Common difference = \[{{T}_{2}}-{{T}_{1}}=3-1=2\]
Now, let us denote the first term of the A.P. with ‘a’ common difference with ‘d’. Therefore, we have,
\[\Rightarrow \] a = 1 and d = 2
We know that the sum of ‘n’ terms of an A.P. is given as: -
\[\Rightarrow \] \[{{S}_{n}}=\dfrac{n}{2}\left[ 2a+\left( n-1 \right)d \right]\], where ‘\[{{S}_{n}}\]’ is the required sum.
So, substituting the values of ‘a’ and ‘d’, we get,
\[\begin{align}
  & \Rightarrow {{S}_{n}}=\dfrac{n}{2}\left[ 2\times 1+\left( n-1 \right)\times 2 \right] \\
 & \Rightarrow {{S}_{n}}=\dfrac{n}{2}\left[ 2+2n-2 \right] \\
 & \Rightarrow {{S}_{n}}=\dfrac{n}{2}\times 2n \\
 & \Rightarrow {{S}_{n}}={{n}^{2}} \\
\end{align}\]
Hence, option (a) is the correct answer.

Note: One may note that the given series is the sum of first ‘n’ odd natural numbers whose sum is given as a formula: - \[{{S}_{n}}={{n}^{2}}\]. So, we can directly remember the result and answer the question without solving it. One may use another formula given as: - \[{{S}_{n}}=\dfrac{n}{2}\left[ {{T}_{1}}+{{T}_{n}} \right]\] to get the answer. Here, \[{{T}_{1}}\] is the first term and \[{{T}_{n}}\] is the \[{{n}^{th}}\] term of the A.P. Here, in this question the most important thing is to determine the type of series because then only we can apply the sum of n terms formula.