Sum of certain consecutive odd positive numbers is ${{57}^{2}}-{{13}^{2}}$ . Find them.
Answer
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Hint: The question is related to the sum of consecutive odd positive numbers. The sum of first $n$ consecutive odd positive numbers is equal to ${{n}^{2}}$ . So, the given expression represents the sum of odd numbers greater than $13$ and less than or equal to $57$ .
Complete step-by-step answer:
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 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 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 first $n$ consecutive odd positive numbers is equal to ${{n}^{2}}$ .
Now, coming to the question, we are given the expression ${{57}^{2}}-{{13}^{2}}$. If we analyze the expression carefully, we can conclude that it represents the difference of the sum of first $57$ consecutive odd positive numbers and first $13$ consecutive odd positive numbers. So, the expression ${{57}^{2}}-{{13}^{2}}$ represents the sum of odd numbers greater than $13$ and less than or equal to $57$ . So, the numbers are $15,17,19,21,23,25,27,29,31,33,35,37,39,41,43,45,47,49,51,53,55$ and $57$.
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:
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 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 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 first $n$ consecutive odd positive numbers is equal to ${{n}^{2}}$ .
Now, coming to the question, we are given the expression ${{57}^{2}}-{{13}^{2}}$. If we analyze the expression carefully, we can conclude that it represents the difference of the sum of first $57$ consecutive odd positive numbers and first $13$ consecutive odd positive numbers. So, the expression ${{57}^{2}}-{{13}^{2}}$ represents the sum of odd numbers greater than $13$ and less than or equal to $57$ . So, the numbers are $15,17,19,21,23,25,27,29,31,33,35,37,39,41,43,45,47,49,51,53,55$ and $57$.
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: 22nd Sep 2023
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