The sum of the first $20$ odd natural numbers is:
A.$100$
B.$210$
C.$400$
D.$420$
Answer
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Hint: The question is related to the sum of consecutive odd positive numbers. The sum of the first $n$ consecutive odd positive numbers is equal to ${{n}^{2}}$ .
Complete step-by-step answer:
Before we proceed with the solution, we must understand the concept of an arithmetic progression. An arithmetic progression is a series of numbers, in which the difference between any two consecutive numbers is constant. In the series of odd numbers, we can see that the difference between any two consecutive odd numbers is always equal to $2$ . For example: Consider the series $5,7,9,11,13$ . It is a series of odd natural numbers with the first term $5$ and the last term $13$. Now, the difference between the first two terms is $7-5=2$ , the difference between the next two terms is $9-7=2$ , the difference between the next two terms is $11-9=2$ , and the difference between the last two terms is $13-11=2$ . So, we can say that the series of consecutive odd numbers is an arithmetic progression with common difference $2$ . If the first $n$ odd numbers are considered then, it is an arithmetic progression with the first term equal to $1$ , common difference equal to $2$ , and the number of terms equal to $n$ . Now, we know, the sum of the first $n$ terms of an arithmetic progression with the firth term equal to $a$ and the common difference equal to $d$ is given as ${{S}_{n}}=\dfrac{n}{2}\left[ 2a+\left( n-1 \right)d \right]$ . So, the sum of the first $n$ consecutive odd positive numbers will be equal to $\dfrac{n}{2}\left[ 2+\left( n-1 \right)2 \right]$ .
\[=\dfrac{n}{2}\left[ 2+2n-2 \right]\]
\[=\dfrac{n}{2}\left[ 2n \right]\]
$={{n}^{2}}$
So, the sum of the first $n$ consecutive odd positive numbers is equal to ${{n}^{2}}$ .
Now, coming to the question, we are asked to find the sum of first $20$ odd natural numbers. So, here $n=20$ . So, the sum is given as ${{20}^{2}}=400$ . Hence, the sum of the first $20$ odd natural numbers is equal to $400$ .
So, option C. is the correct answer.
Note: The sum of first $n$ natural numbers is equal to $\dfrac{n\left( n+1 \right)}{2}$ . The sum of first $n$ even natural numbers is equal to $\dfrac{n\left( n+2 \right)}{2}$ and the sum of first $n$ odd natural numbers is equal to ${{n}^{2}}$ . These formulae should be remembered as they are frequently used and there should be no confusion between the formulae.
Complete step-by-step answer:
Before we proceed with the solution, we must understand the concept of an arithmetic progression. An arithmetic progression is a series of numbers, in which the difference between any two consecutive numbers is constant. In the series of odd numbers, we can see that the difference between any two consecutive odd numbers is always equal to $2$ . For example: Consider the series $5,7,9,11,13$ . It is a series of odd natural numbers with the first term $5$ and the last term $13$. Now, the difference between the first two terms is $7-5=2$ , the difference between the next two terms is $9-7=2$ , the difference between the next two terms is $11-9=2$ , and the difference between the last two terms is $13-11=2$ . So, we can say that the series of consecutive odd numbers is an arithmetic progression with common difference $2$ . If the first $n$ odd numbers are considered then, it is an arithmetic progression with the first term equal to $1$ , common difference equal to $2$ , and the number of terms equal to $n$ . Now, we know, the sum of the first $n$ terms of an arithmetic progression with the firth term equal to $a$ and the common difference equal to $d$ is given as ${{S}_{n}}=\dfrac{n}{2}\left[ 2a+\left( n-1 \right)d \right]$ . So, the sum of the first $n$ consecutive odd positive numbers will be equal to $\dfrac{n}{2}\left[ 2+\left( n-1 \right)2 \right]$ .
\[=\dfrac{n}{2}\left[ 2+2n-2 \right]\]
\[=\dfrac{n}{2}\left[ 2n \right]\]
$={{n}^{2}}$
So, the sum of the first $n$ consecutive odd positive numbers is equal to ${{n}^{2}}$ .
Now, coming to the question, we are asked to find the sum of first $20$ odd natural numbers. So, here $n=20$ . So, the sum is given as ${{20}^{2}}=400$ . Hence, the sum of the first $20$ odd natural numbers is equal to $400$ .
So, option C. is the correct answer.
Note: The sum of first $n$ natural numbers is equal to $\dfrac{n\left( n+1 \right)}{2}$ . The sum of first $n$ even natural numbers is equal to $\dfrac{n\left( n+2 \right)}{2}$ and the sum of first $n$ odd natural numbers is equal to ${{n}^{2}}$ . These formulae should be remembered as they are frequently used and there should be no confusion between the formulae.
Last updated date: 21st Sep 2023
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