Find the value of $n$.

Question

Write the value of $n$ if the sum of n terms of the series $1+3+5+7...n =n^2$.

I'm not getting the right value if I proceed with the general formula for finding sum of n terms of a arithmetic series. The general summation formula for arithmetic series is $\frac{n(2a+(n-1)d)}{2}$, where $a$ is the first term, $d$ is the common difference and $n$ is the number of terms.

Answer 1

Note that the sum of an arithmetic progression with $n$ terms having a starting term $a$ and common difference $d$ is: $$S_n = \frac{n}2 \left[2a + (n-1)d\right]$$ Now, if $a=1, d=2$, what do you get?

$$S_n = \frac{n}{2}\left[2+(n-1)2\right] = \frac{n}{2}\left[2n \right] = n^2$$

Answer 2

If we add $2+4+...+(n-1) = 2(1+2+3+...+\frac{n-1}{2})$ to both sides of the equation, we get $$1+2+3+...+n = n^2+2\bigg(1+2+3+...+\frac{n-1}{2}\bigg)$$ which can be written as: $$\frac{n(n+1)}{2} = n^2+2\bigg[\frac{\big(\frac{n-1}{2}\big)\big(\frac{n+1}{2}\big)}{2} \bigg]$$ that is $$\frac{n^2+n}{2} = n^2+\frac{n^2-1}{4} \implies n^2+n = \frac{5n^2-1}{2} \implies 3n^2-2n-1 = 0 \implies n = 1$$ because $n$ is not a negative number.

Answer 3

Since we have $$1+3+5+7+\cdots (2n-1)=n^2$$ and we want to have $$1+3+5+7+\cdots +n=n^2$$ we must have $2n-1=n$ , which implies $n=1$