# help for proving an equation by induction

For this equation:

$$-1^3+(-3)^3+(-5)^3+\ldots+(-2n-1)^3=(-n-1)^2(-2n^2-4n-1)$$

how can I prove this by induction?

When I set $n = 1$ for the base case I got:

$$-1^3 + (-3)^3 + (-5)^3 + \ldots + (-3)^3 = -28$$

but am having trouble with the following inductive steps

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Please, it hurts me to see all these minus signs. The first thing to do is to get rid of them. Our question is equivalent to showing that $1^3+\cdots+(2n+1)^3=(n+1)^2(2n^2+4n+1)$. I am allergic to minus signs, but am not asking just for me. You have a much better chance of succeeding if you get rid of them. – André Nicolas Feb 23 '13 at 23:11

You’d have had an even simpler base case had you started with $n=0$: $(-1)^3=(-1)^2(-1)$.

For the induction step your induction hypothesis should be that the result is true for some particular $n$, i.e., that

$$(-1)^3+(-3)^3+(-5)^3+\ldots+(-2n-1)^3=(-n-1)^2(-2n^2-4n-1)\;,\tag{1}$$

and you’ll try to prove the corresponding statement about $n+1$. The first step is to figure out what that statement is:

\begin{align*} (-1)^3+(-3)^3&+\ldots+(-2n-1)^3+\big(-2(n+1)-1\big)^3\\ &=\big(-(n+1)-1\big)^2\big(-2(n+1)^2-4(n+1)-1\big) \end{align*}

or, after a bit of algebraic simplification,

\begin{align*} (-1)^3+(-3)^3+\ldots+(-2n-1)^3+\big(-2n-3\big)^3&=(-n-2)^2(-2n^2-8n-7)\\ &=-(n+2)^2(2n^2+8n+7)\;. \end{align*}\tag{2}

The lefthand side of $(2)$ can be split into two pieces as

$$\Big((-1)^3+(-3)^3+\ldots+(-2n-1)^3\Big)+\big(-2n-3\big)^3\;,$$

and the induction hypothesis $(1)$ tells us that the first piece is $$(-2n-1)^3=(-n-1)^2(-2n^2-4n-1)=-(n+1)^2(2n^2+4n+1)\;.$$

Thus,

\begin{align*} (-1)^3+(-3)^3&+\ldots+(-2n-1)^3+\big(-2n-3\big)^3\\ &=-(n+1)^2(2n^2+4n+1)+(-2n-3)^3\;, \end{align*}

and all that you have to do now in order to prove $(2)$ (and thereby complete the induction step) is show that

$$-(n+1)^2(2n^2+4n+1)+(-2n-3)^3=-(n+2)^2(2n^2+8n+7)\;.$$

This is just algebra, and I’ll leave it to you.

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Inductive proofs of sums like yours are easily tackled using a very simple general method known as telescopy - see the trivial inductive proof here, of the following fundamental

Theorem $\rm \displaystyle\ \ \sum_{i\,=\,0}^n\, f(i)\, =\, g(n)\iff f(0) = g(0)\ {\rm\ and\ }\ f(n) \,=\, g(n)-g(n\!-\!1)\:\$ for $\rm\,n \ge 1.$

$$\rm f(n) \,=\, (2n+1)^3,\quad g(n) \,=\, (n+1)^2 (2n^2+4n+1)$$

Thus, applying the theorem, we check that $\rm\ f(0) = 1 = g(0)\$ and it remains to check that $\rm\, g(n)-g(n-1) = f(n)\:$ for $\rm\:n\ge 1,\:$ which is rote arithmetic. Notice that the proof requires no ingenuity at all - only verifying some simple polynomial equalities - a purely mechanical process. You can find many more examples of telescopy and related results in other answers here.

Remark $\$ The other answers all end up verifying the same equality $\rm\, g(n)-g(n-1) = f(n).\:$ This is no coincidence. They are effectively repeating the same linked telescopic sum proof for a special case of $\rm\:f(n)\:$ and $\rm\:g(n).\:$ No insight is gained by repeating such telescopy proofs ad infinitum for special cases, getting hopelessly lost in messy algebraic calculations intrinsic in motley special cases (details that greatly obfuscate the simple telescopic inductive structure). Doing so would be akin to reproving the fundamental theorem of calculus for each specific function it is applied to (the above is a discrete analog - the fundamental theorem of difference calculus).

Much more inductive insight is gained by proving once and for all the above general telescoping sum formula. Here, rid of all the messy details of special cases, one sees clearly the beautiful way that telescopic induction works, and this lends more insight on induction in general. Such telescopic induction skills will prove quite handy later on, since many inductive proofs of sums and products have this special telescopic form. Being able to quickly dispatch these trivial inductions using these tools will help you to not lose focus on the essential aspects of the problem you are solving (which for non-logicians are rarely of inductive nature).

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Very nice argument. – André Nicolas Feb 24 '13 at 4:14

First, as the power of each summand is odd, I'd write the equation as

$$-1^3-3^3-5^3-\ldots-(2n+1)^3=-(n+1)^2(2n^2+4n+1)\Longleftrightarrow$$

$$1^3+3^3+5^3+\ldots+(2n+1)^3=(n+1)^2(2n^2+4n+1)$$

The base case is

$$n=1:\;\;\;\;\;1^3+3^3=28\stackrel ?=(1+1)^2(2+4+1)=28\ldots\ldots good$$

Assume for $\,n\,$ and show for $\,n+1\,$:

$$1^3+3^3+\ldots+(2n+1)^3+(2n+3)^3\stackrel{\text{Ind. hypothesis}}=(n+1)^2(2n^2+4n+1)+(2n+3)^3=\ldots$$

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