Find seventh term of the evaluation of $(2a+b^2)^8$ I need to find the seventh term of 
$(2a+b^2)^8$ 
But, at this moment, the only way to solve this I can make up is to distribute the brackets.. 
Is there any other way to do this?

See here.
Using the binomial theorem, we can expand $(a+b)^8$ to be
${8 \choose 0}a^8 + {8 \choose 1}a^7b + {8 \choose 2}a^6b^2 + {8 \choose 3}a^5b^3 + {8 \choose 4}a^4b^4 + {8 \choose 5}a^3b^5 + {8 \choose 6}a^2b^6+{8 \choose 7}ab^7 + {8 \choose 8}b^8$
From this we get the seventh term to be ${8 \choose 6}a^2b^6$ or $28a^2b^6$.
Note that in the above expansion, we used a and b. To fit your formula of $(2a + b^2)^8$ we simply replace a and b for the answer of ${8 \choose 6}(2a)^2(b^2)^6$ or $7168a^2b^{12}$.
