Question #d17b5

1 Answer
Mar 11, 2017

Answer:

The Reqd. both Co-effs. are #""_12C_4=495=_12C_8.#

Explanation:

The General #(r+1)^(th)# Term, #T_(r=1)# in the Expansion of

#(a+b)^n# is, #T_(r+1)=""_nC_ra^(n-r)b^r, r=0,1,2,...,n.#

Since, #n=12, T_(r+1)=""_12C_ra^(12-r)b^r, r=0,1,2,...,12......(star).#

To find the co-eff. of #a^8b^4#, we compare this with, #(star).#

We find that,#12-r=8,r=4; i.e., r=4.#

This means that #a^8b^4# is #T_(r+1)=T_5# in #(a+b)^12.#

Hence, the co-eff. of #a^8b^4# is #""_12C_4=(12!)/{(4!)(12-4)!}#

#=(12!)/{(4!)(8!)}={cancel(12)(11)(10)(9)cancel(8!)}/{(1)(2)cancel(3)cancel(4)cancel(8!)}=(11)(5)(9)=495.#

Similarly, we can find the co-eff. of #a^4b^8,# which is,

#""_12C_8=""_12C_4=495.#

The later follows from the fact that,

#""_nC_r""=""_nC_(n-r)"".#

Enjoy Maths.!