Given that $\sum_{a=1}^{b}a=\frac{a(a+1)}{6}$ prove through induction that $$\sum_{a=1}^{b}a(b-a)=\frac{b(b-1)(b+6)}{6}$$ Normally I would start by showing that this statement is true for $b=1$ and move on to show that this statement is true for $b=a+1$, but I'm having trouble in this because in my previous induction proves there was only $b$ or only $a$ but now there is $a$ and $b$.
What is the method here?
On
For an induction, of the title formula, get rid of the summation index $a$, by rewriting $$ 1\cdot(b-1)+2(b-2)+3(b-3)+\cdots (b-1)\cdot 1=f(b) $$ with the correct poylnomial $f(b)$. Start with $b=1$. What do you obtain for $b\mapsto b+1$?
On
As it is written your proposition is not true.
The left side is clearly an integer, while for $b=2$ the right side produces a fraction. There is no factor divisible by $3$ in the numerator.
However, when you have it sorted out what proposition you are actually trying to prove...
Take care of the base case.
Assume:
$\sum_\limits{a = 1}^{b} a(b-a) = f(b)$
Show.
$\sum_\limits{a = 1}^{b+1} a(b+1-a) = f(b+1)\\ \sum_\limits{a = 1}^{b+1} a(b+1-a) = \sum_\limits{a = 1}^{b} a(b+1-a)$
Because the last term in the series on the $LHS = 0.$
$\sum_\limits{a = 1}^{b} a(b+1-a)\\ \sum_\limits{a = 1}^{b} (a(b-a) + a)\\ f(b) + \sum_\limits{a = 1}^{b} a$
By the inductive hypothesis.
You have probably already proven that $\sum_\limits{a = 1}^{b} a = \frac 12 b(b+1)$
And then you will need to do just a little algebra to show that:
$f(b) + \frac 12 b(b+1) = f(b+1)$
Notice that $a(b-a)=ab-a^2$. Then the sum $S$ becomes:
$$ S=\sum_{a=1}^{b}a(b-a)=\sum_{a=1}^{b}ab-\sum_{a=1}^{b}a^2=b\sum_{a=1}^{b}a-\sum_{a=1}^{b}a^2 $$
By hypothesis $\displaystyle\sum_{a=1}^{b}a=\frac{b(b+1)}{6}$. By induction you can prove that $\displaystyle\sum_{a=1}^{b}a^2 = \frac{1}{6}(2b+1)b(b+1)$. Finally:
$$ b\left(\frac{b(b+1)}{6}\right)-\frac{1}{6}(2b+1)b(b+1) = \frac{-1}{6}b(b+1)^2 $$