Does someone know a shortcut for finding the coffiecient of the product terms in the binomial expansion?? eg: Cofficient of x^5 in the expansion of ((1+x)^6 (3-4x)^8) ??

Question

ganeshie8 · Answer

im not sure of any shortcuts, but u could simplify the problem a little bit... you only need to consider :
(1+x)^5(3-4x)^5

ganeshie8 · Answer

5 = 0+5 = 1+4 = 2+3 = 3+2 = 4 + 1 = 5+ 0

ganeshie8 · Answer

so that gives 6 additions which is really not what you're looking for

anonymous · Answer

kk

anonymous · Answer

This is the general way ...anyways thank you!!

ganeshie8 · Answer

you familiar wid below thing right ?

$$^nC_0 + ^n C_1 + ^nC_2 +... + ^nC_n = 2^n$$

ganeshie8 · Answer

we may use it and derive some formula i think

anonymous · Answer

kk