Question #d17b5

Question

1298 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-03-11T15:12:38

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

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.!