number of terms The following  problem maybe tedious if done by hand and requires patience. After factorizing the following variables find the number of terms and the sum of the number of terms.
$(a^0),(a+b)^0,(a+b+c)^0,(a+b+c+d)^0,(a+b+c+d+e)^0,(a+b+c+d+e+f)^0$
$(a^1),(a+b)^1,(a+b+c)^1,(a+b+c+d)^1,(a+b+c+d+e)^1,(a+b+c+d+e+f)^1$
$(a^2),(a+b)^2,(a+b+c)^2,(a+b+c+d)^2,(a+b+c+d+e)^2,(a+b+c+d+e+f)^2$
$(a^3),(a+b)^3,(a+b+c)^3,(a+b+c+d)^3,(a+b+c+d+e)^3,(a+b+c+d+e+f)^3$
$(a^4),(a+b)^4,(a+b+c)^4,(a+b+c+d)^4,(a+b+c+d+e)^4,(a+b+c+d+e+f)^4$
$(a^5),(a+b)^5,(a+b+c)^5,(a+b+c+d)^5,(a+b+c+d+e)^5,(a+b+c+d+e+f)^5$
$(a^6),(a+b)^6,(a+b+c)^6,(a+b+c+d)^6,(a+b+c+d+e)^6,(a+b+c+d+e+f)^6$

What you are searching for is the multinomial theorem. 
The formula is
$$ (x_1+x_2+...+x_m)^n = \sum_{k_1+...+k_m=n} {n \choose k_1,...,k_m} x_1^{k_1}...x_m^{k_n} $$
where 
$$ {n \choose k_1,...,k_m} = \frac{n!}{k_1!k_2!...k_m!}.$$
For a proof take a look at the Wikipedia post: http://en.wikipedia.org/wiki/Multinomial_theorem
