What is the closed summation form for $1 + 3 + 5 + \cdots + (2n-1)$ ?
I know that the closed form for $1 + 2 + 3+\cdots + n = n(n+1)/2$ and I tried plugging in $(2n-1)$ for $n$ in that expression, but it didn't produce a correct result:
$(2n-1)((2n-1)+1)/2$
plug in 3
$(2n-1)((2n-1)+1)/2 = 6 != 1 + 2 + 3 = 9$
HINT : $$1+3+5+\cdots+(2n-1)=\sum_{k=1}^{n}(2k-1)=2\sum_{k=1}^{n}k-\sum_{k=1}^{n}1.$$
Sum of Arithmetic Progression $=\frac{N}{2}(a_1+a_n)$, where N is the nunber of terms
$N = \frac{(a_n-a_1)}{d} + 1 = \frac{(2n-2)}{2} + 1 = n-1+1 = n$
Thus the sum is $ = \frac{n}{2}(1+2n-1) = n^2$
$$M=1+3+5+7+...+(2n-3)+(2n-1)\\M=(2n-1)+(2n-3)+...+7+5+3+1\\M+M=(1+2n-1)+(3+2n-3)+(5+2n-5)+....(2n-3+3)+(2n-1+1)\\n-term\\M+M=n(2n)\\2M=2n^2\\M=n^2 $$
Let $$S_n=1+2+3+...+(2n-1)$$
Clearly
$S_1=1=1^2$
$S_2=1+3=2^2$
$S_3=1+3+5=9=3^2$
Suppose $S_n=n^2,$ then $S_{n+1}=S_n+(2n+1)=(n+1)^2$
By mathematical induction $$S_n=n^2 ,\forall n\in \mathbb{N}$$
If you write down the first five elements of the sequence, they are 1, 4, 9, 16 and 25. That should be enough for a guess which you can prove using induction.
