Question

Find the greatest term in the expansion of (1+2x)^9, where x = 2/3.

Answer 1

Answer:

Is this your answer

Step-by-step explanation:

The nth term of (32+203)12 is tn=(12n)(32)12−n(203)n for n=0,1,…,12. The ratio of consecutive terms is ρn=tn+1tn=409(12−nn+1).

These ratios are easy to compute

nrn01603122092400273104649514027680217259816081943108099111027

The last occurence of rn>1 happens at n=9. So t10=(1210)(32)2(203)10 will be the largest term.

As @BarryCipra has pointed out. It is not necessary to compute the above table. We can solve the following inequality.

tn+1tn409(12−nn+1)480−40n49nnn≥1≥1≥9n+9≤472≤47249≤9