# Expanding Equation with Binomial Theorem

How do I expand this equation: $(1+t+t^2)^5$

I formed the equation into a binomial equation this way: $(1+t+t^2)^5=\sum \binom{5}{r_1}\binom{5-r_1}{r_2}t^{r_2}t^{2r_1}$

But I cannot remember how to continue from here to solve for the $r_1$ and $r_2$ terms from here and then to further expand it.

Thanks for any help!

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You'll actually need to use multinomials here.... – J. M. Aug 21 '11 at 17:30
But how do I find out the $r_1$ and $r_2$ before I could carry out the multinomial expansion? Actually, this is the first time I'm seeing this multinomial. I formed into this binomial equation by splitting the $t$ up into two parts and then further split them, which is why now I have these 2 unknowns $r_1$ and $r_2$. – xenon Aug 21 '11 at 17:34
@xEnOn: You can do it this way by setting the lower bound of $r_2$ to zero and its upper bound to $5-r_1$. Then the bounds for $r_1$ are 0 and 5. After all, in order for both binomial coefficients to be defined you need $0\le r_2\le 5-r_1$ and $0\le r_1\le 5$. The bounds on the inner sum will necessarily depend on the summation variable of the outer sum. – Jyrki Lahtonen Aug 21 '11 at 18:11
If you were intended to exclusively use binomial coefficients, @Jyrki's is the way to go; on the other hand, going through that route has been confusing for me, and I use what yunone wrote in his answer. – J. M. Aug 21 '11 at 18:50
Hint: $$(1+t+t^2)^5=(1+[t+t^2])^5$$ – Listing Aug 21 '11 at 19:58

You could use the trinomial expansion, a specific case of the multinomial formula. Your formula would then be $$(1+t+t^2)^5=\sum_{i,j,k}\binom{n}{i,j,k}1^it^j(t^2)^k$$ for $i,j,k$ nonnegative, where $i+j+k=5$. It's then just a matter of finding all possible sums and plugging in.

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You can take $(1+t)$ as $a$ and $t^2$ as $b$. Then solve by using same binomial formula.

$$1+5t+15t^2+30t^3+45t^4+51t^5+45t^6+30t^7+15t^8+5t^9+t^{10}$$