Question

Sum of $n$ odd number of consecutive numbers is divisible by $n$ . Explain the reason.

Answer 1

Hint: To prove the given statement we will have to first find the sum of consecutive odd numbers and then we will first start adding the first odd number, then first two and then first three odd numbers and then generalise it. After that, we will see that the number of digits added collectively is always equal to the square root of the total number and we will again prove that by first testing samples and then generalise it.

Complete step-by-step answer:
First, we will find the sum of odd consecutive numbers.
For that, we need to understand the pattern of odd numbers sequence to prove their sum:
\[\Rightarrow \] Sum of first odd number =\[1\]
\[\Rightarrow \] Sum of first two odd numbers = $1+3=4\left( 4=2\times 2={{2}^{2}} \right)$
\[\Rightarrow \] Sum of first three odd numbers = $1+3+5=9\left( 9=3\times 3={{3}^{2}} \right)$
\[\Rightarrow \] Sum of first four odd numbers = $1+3+5+7=16\left( 16=4\times 4={{4}^{2}} \right)$

The number of digits added collectively is always equal to the square root of the total number that is:
Sum of first odd number =\[1\]. The square root of $1,\sqrt{1}=1$ so, only one digit was added.
Sum of the first two odd numbers =\[4\]. The square root of $4,\sqrt{4}=2$ so, two digits were added.
Sum of the first three odd numbers =\[9\]. The square root of $9,\sqrt{9}=3$ so, three digits were added.
Sum of the first four odd numbers =\[16\]. The square root of $16,\sqrt{16}=4$ so, four digits were added.

Hence, from the above estimation, we can say that the formula to find the sum of the first $n$ odd numbers is $n\times n={{n}^{2}}$.
For example, if we put$n=22$ , then we have $n\times n={{n}^{2}}=22\times 22=484$, which is equal to the sum of the first $22$ odd numbers.
So, we see that the sum of first $n$ odd numbers is ${{n}^{2}}$ and it is divisible by $n$ .

Note: If we don’t know the number of odd numbers present in a series, then the formula to determine the sum between $1$ and $n$ is ${{\left( \dfrac{1}{2\left( n+1 \right)} \right)}^{2}}$ . In such questions you should always give an example after providing the generalised formula.