Question

Find the 6th term of the expansion (3x²-1/3x)8

Answer 1

Answer:

Step-by-step explanation:

First of all, you do realise that your expression reads as

(32x−13x)11

which can be rearranged into

(9−26x)11

(76x)11

711611x11.

Let’s assume you have something of the form

(ax+by)d

where a,b,d are integers. Then to expand this you need to use binomial expansion which it can be written in the form

(ax+by)d=∑i=0d(ax)d−i(by)i(dd−i)

For example if d=3 we get

(ax+by)3=(ax)3(by)0(33−0)+(ax)3−1(by)1(33−1)+(ax)3−2(by)2(33−2)+(ax)3−3(by)3(33−3)

(ax+by)3=(ax)3(1)(33)+(ax)2(by)(32)+(ax)(by)2(31)+(1)(by)3(30)

(ax+by)3=(ax)3+(ax)2(by)3+(ax)(by)23+(by)3

where we have use the formula for the binomial coefficients

(Nk)=N!k!(N−k)!

So if d=11 and we are looking for the 6th term from the end, you are looking for the 7th term from the start, since the binomial expansion has d+1 terms, that is (11+1)−6=6 plus 1 because we are counting from zero, thus the 7th term. In other words we are looking for the i=6 term in our sum series. So the binomial coefficient will be

(116)=11!6!(11−6)!=11×10×9×8×75!=11×10×9×8×75×4×3×2=11×2×3×7=462.

If a=32 and b=−13 then we have

(32)11−6(−13)6462

3525(−1)6136462

4623×25=4623×32=46296=11×2×3×73×2×24=7716

If in your expresion x and y appear as1x and 1y,it bears no difference, the term will simply be

462481x51y6=77161x51y6