Question

Express 64 as the sum of 8 odd numbers.

Answer 1

Hint: Consider the general form of the odd terms as ${{T}_{n}}=2n-1$ where n is the ${{n}^{th}}$ odd term. Now, find the general formula for the sum of first n odd natural numbers using the formula ${{S}_{n}}=\dfrac{n}{2}\left[ {{T}_{1}}+{{T}_{n}} \right]$. Here find the value of ${{T}_{1}}$ by substituting 1 in the formula ${{T}_{n}}=2n-1$. Once the general formula for the sum of n odd numbers is found, substitute it with 64 and solve for the value of n. Start with n = 1 and find the values of ${{T}_{n}}$ for each n up to the value of n obtained above.

Complete step by step solution:
Here we have been asked to write 64 as the sum of 8 odd numbers. So, first let us find the general formula for the sum of first n odd natural numbers.
Now, we know that the general form of odd numbers is given by the relation ${{T}_{n}}=2n-1$ where n denotes the ${{n}^{th}}$ odd numbers. Clearly we can see that the odd successive odd terms will form an A.P. with common difference as 2 and first term as 1. So the sum of the n terms of this A.P. will be given by the formula ${{S}_{n}}=\dfrac{n}{2}\left[ {{T}_{1}}+{{T}_{n}} \right]$.
So we have, first term = ${{T}_{1}}=2\left( 1 \right)-1=1$.
$\begin{align}
  & \Rightarrow {{S}_{n}}=\dfrac{n}{2}\left[ 1+2n-1 \right] \\
 & \Rightarrow {{S}_{n}}=\dfrac{n}{2}\left[ 2n \right] \\
 & \Rightarrow {{S}_{n}}={{n}^{2}} \\
\end{align}$
Now, the above sum should be equal to 64, so we have,
$\Rightarrow {{n}^{2}}=64$
Taking square root both the sides we get,
$\Rightarrow n=8$
The above value n = 8 means we have to start with n = 1 and find the values of ${{T}_{n}}$ up to n = 8. Therefore, eight odd numbers will be 1, 3, 5, 7, 9, 11, 13 and 15. Hence we can write:
$\Rightarrow $ 64 = 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15

Note: Initially we didn’t know how to start with which odd number but after finding the value n = 8 it was clear that we have to take the first 8 odd positive numbers. You must remember the formula of sum of first n odd natural numbers as well as the sum of first n even natural numbers which is given as $\dfrac{n}{2}\left( n+1 \right)$.