What is the numerically greatest or magnitude wise greatest term in the expansion?

Question

I am just a beginner to Binomial Theorem and I want to find out the numerically or magnitude wise greatest term in the expansion of $(3-5x)^{11} $ when $ x=\frac{1}{5} $

I know some methods, but they are quite confusing. And I want a simple method. So if anyone has any idea they are welcomed.

Answer 1

Note that the general term of the expression $(3-5x)^{11} $ is $$T_r = \binom {11}{r}3^r (-5x)^{11-r} $$ When $T_r $ is the largest term then: $$\frac {T_{r+1}}{T_r} \leq 1$$ $$\implies \frac {\binom {11}{r+1}}{\binom {11}{r}} \times 3 \times (-5x)^{-1} \bigg \lvert_{ x = \frac15} \leq 1$$

Ignore the negative sign as we want the greatest numerical value.

Surely you can take it from here.

Answer 2

When $x = \frac15$, we just have $(3-1)^{11}$. Then the question is basically "what is the largest $n$ such that $3\binom{11}{n-1}\leq\binom{11}{n}$", since the change from one term to the next is that the "denominator" of the binomial coefficient increases by $1$, and the exponent of $3$ decreases by $1$, and we want to know at what point the change in the binomial coefficient doesn't outweigh the change in the exponent of $3$ (ignoring sign).

(Alternatively, one may derive this inequality from "one term is larger than the previous if their ratio is larger than $1$" as the other answer does. Arguably, that's easier to follow than the above paragraph.)

We have that $$ 3\binom{11}{n-1}\leq\binom{11}{n}\\ 3\leq\frac{\binom{11}{n}}{\binom{11}{n-1}}\\ 3\leq\cfrac{\frac{11!}{(n)!(11-n)!}}{\frac{11!}{(n-1)!(11-n+1)!}}\\ 3 \leq \frac{11-n+1}{n}\\ 3n\leq12-n\\ 4n\leq12\\ n\leq3 $$ So we actually get that $n = 3$ is the largest $n$ for which the previous term isn't larger (again, ignoring sign), and that term is $$ 3^{8}(-1)^3\binom{11}3 = -1\,082\,565 $$ Since the inequality gave $n\leq$ an integer, the previous term is actually equal in absolute value, and has positive sign, so numerically, that's the largest. In general, if one term is negative, but in absolute value the strictly largest (i.e. the inequality you get from the above argument is $n\leq$ some non-integer), then you will have to compare the term before and the term after to see which one is numerically larger, but it will necessarily be one of those two.