Question

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) ??

Answer 1

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

Answer 2

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

Answer 3

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

Answer 4

kk

Answer 5

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

Answer 6

you familiar wid below thing right ?

\[^nC_0 + ^n C_1 + ^nC_2 +... + ^nC_n = 2^n\]

Answer 7

we may use it and derive some formula i think

Answer 8

kk