Question

The value of $1 + 3 + 5 + $- - - - - - - - - - - - - - - - - -$ + 25$ is
(A) $196$
(B) $625$
(C) $225$
(D) $169$

Answer 1

Hint: The given series is an arithmetic series with the common difference $2$ and we have to find the summation of this arithmetic series. Firstly, find the number of terms in this series and then apply the formula for the summation of arithmetic progression.

Formula used: ${a_n} = a + \left( {n - 1} \right)d$
${S_n} = \dfrac{n}{2}\left\{ {2a + \left( {n - 1} \right)d} \right\}$ where ${a_n}$ is ${n^{th}}$ term of the AP, $a$ is the first term, $n$ is the number of terms, $d$ is the common difference and ${S_n}$ is the sum of $n$ terms of Arithmetic progression.

Complete step-by-step solution:
The given series is $1 + 3 + 5 + $- - - - - - - - - - - - - - - - - -$ + 25$.
Here, first term $a = 1$, common difference $d = 3 - 1 = 2$, ${n^{th}}$ term ${a_n} = 25$
Now, apply the formula for the $n$ term to get the value of $n$. And putting the value of ${a_n},a$ and $d$we get,
$
   \Rightarrow {a_n} = a + \left( {n - 1} \right)d \\
   \Rightarrow 25 = 1 + \left( {n - 1} \right)2 \\
   \Rightarrow 25 - 1 = \left( {n - 1} \right)2 \\
   \Rightarrow \dfrac{{24}}{2} = \left( {n - 1} \right) \\
   \Rightarrow \left( {n - 1} \right) = 12 \\
   \Rightarrow n = 12 + 1 \\
  \therefore n = 13
 $
So, the number of terms in the given series is $13$.
Now, we have to find the summation of AP that can be calculated by using the above given formula. ${S_n} = \dfrac{n}{2}\left\{ {2a + \left( {n - 1} \right)d} \right\}$
by putting the value of $a,n$ and $d$. We get,
$
   \Rightarrow {S_n} = \dfrac{{13}}{2}\left\{ {2 \times 1 + \left( {13 - 1} \right)2} \right\} \\
   \Rightarrow {S_n} = \dfrac{{13}}{2}\left\{ {2 + 12 \times 2} \right\} \\
   \Rightarrow {S_n} = \dfrac{{13}}{2}\left\{ {2 + 24} \right\} \\
   \Rightarrow {S_n} = \dfrac{{13}}{2} \times 26 \\
   \Rightarrow {S_n} = 13 \times 13 \\
  \therefore {S_n} = 169
 $
Thus, the summation of given series is $169$.

Hence, Option (D) is correct.

Note: This can be alternatively solved by using the formula for the summation of the first $n$ consecutive odd number.
Summation of $n$ consecutive odd number $ = {n^2}$
Similarly, we can also find the summation of first $n$ consecutive even numbers by using formula $n\left( {n + 1} \right)$.
Summation of first $n$ consecutive natural numbers $ = n\left( {\dfrac{{n + 1}}{2}} \right)$.