Question

Without adding, find the sum of the following numbers
$1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17$

Answer 1

Hint: In this question we have given some series of odd numbers and we have to find the sum without adding. So for finding the sum of first \[n\] odd natural number without actually adding so for that we have to know one method that is, if you're given a series where we take the summation of first n odd natural numbers i.e. ($1 + 3 + 5 + 7 + 9 + .....{n^{th}}$ odd numbers),
Then for this first we have to count how many odd numbers we are given to sum up. So from the above series we can easily say that the total number of odd number are\[n\], so we can write,
 ($1 + 3 + 5 + 7 + 9 + .....{n^{th}}$ odd numbers),$ = {n^2}$………………...…(1)
Here we use the formula that the sum of odd terms is equal to the square of no of terms.

Complete step-by-step solution:
Step1: We are given the series is $1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17$. For this first we will count how many odd numbers we have given,
So,
First term$ = 1$
Second term$ = 3$
Third term$ = 5$
Fourth term$ = 7$
Fifth term$ = 9$
Sixth term$ = 11$
Seventh term$ = 13$
Eighth term$ = 15$
Ninth term$ = 17$
So on counting the number in the series we can say that in the above series we are given total $9$terms, or in other words we have to find the sum of first nine odd natural numbers, so by equation(1) we can say $n = 9$
Step2: Sum of the terms is equal to ${\text{(number of terms)}}^2$
$1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17$$ = {\left( 9 \right)^2}$
$1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17$$ = 81$

So, $81$ is our required solution for this series.

Note: To solve this type of question we have to first identify whether the odd numbers are in chronological order and if all the odd numbers are present without any missing then we can apply the formula which we have used in solving this question. If the numbers are not in chronological order then we cannot apply the formula we used above.