Mathematical Induction Question About A Basis Step Use mathematical induction to show that:
$1^3 + 3^3 + 5^3 + ... + (2n + 1)^3 = (n+1)^2(2n^2+4n+1)$ whenever $n$ is a positive integer.
I am able to solve this problem, but the thing that confuses me on this is the Basis Step.
P(1): $(2(1) + 1)^3$ $\neq$ $4(7)$, does this mean that this problem can not be proved using mathematical induction, or am I missing something? Thanks.
 A: Don't start at $n = 1$.  Start at $n=0$.
$P(n = 0): (1)^3  = (0 + 1)\left(2\times0^2 + 4*0 + 1\right) =1$
$P(1): 1 + 3^3 = 2^2\times\left(2 + 4 + 1\right)$
A: Method -1: Mathematical Induction 
Notice, the following steps by Mathematical Induction 


*

*Setting $n=1$, we get 
$$1^3+(2\cdot 1+1)^3=(1+1)^2(2(1)^3+4(1)+1)$$
$$28=28$$
The identity holds good for $n=1$.

*Assume that it holds for $n=k$ then we have 
$$1^3+3^3+5^3+\ldots +(2k+1)^3=(k+1)^2(2k^2+4k+1)$$

*Setting $n=k+1$, we get 
$$1^3+3^3+5^3+\ldots +(2k+1)^3+(2k+3)^3=(k+1+1)^2(2(k+1)^2+4(k+1)+1)$$
$$\implies 1^3+3^3+5^3+\ldots +(2k+1)^3=(k+2)^2(2k^2+8k+7)-(2k+3)^3$$
$$=2k^4+8k^3+11k^2+6k+1$$ $$=(k+1)^2(2k^2+4k+1)$$
Which is true from (2)
Hence, the given identity holds true for all positive integers $n\geq 1$
Method-2
Notice, we have
$$1^3+3^3+5^3+\dots +(2n+1)^3$$ 
$$=(1^3+2^3+3^3+4^3+5^3+\dots +(2n)^3+(2n+1)^3)-(2^3+4^3+6^3+\ldots +(2n)^3)$$ 
$$=(1^3+2^3+3^3+4^3+5^3+\dots +(2n)^3+(2n+1)^3)-8(1^3+2^3+3^3+\ldots +n^3)$$ 
$$=\left(\frac{(2n+1)((2n+1)+1)}{2}\right)^2-8\left(\frac{n(n+1)}{2}\right)^2$$ 
$$=(n+1)^2(2n+1)^2-2n^2(n+1)^2$$ 
$$=(n+1)^2[(2n+1)^2-2n^2]$$ 
$$=(n+1)^2[4n^2+1+4n-2n^2]$$
$$=(n+1)^2(2n^2+4n+1)$$ 
A: The problem is that for $n=1$ your sum is $1^3+3^3$ and you are interpreting it as $3^3$ only. So the LHS for P(1) is $1^3+3^3=28$ and the RHS is $4(7)=28$. 
You should prove P(0) first by the way. The LHS in that case is $1^3=1$ and the RHS is $1(1)=1$.
