Maximum coefficient in the expansion of $(5+3x)^{10}$

Though I know that I could simply just expand $$(5+3x)^{10}$$ with the binomial theorem for each power of x, is there a simpler and quicker method of finding out the largest coefficient? After manual expansion, I know that it wouldn't be the coefficient of $$x^6$$ since there are other coefficients that are larger, but how do I prove that there's a quick way to get to the largest coefficient possible?

Write $$(5+3x)^{10}=\sum_{i=0}^{10}a_ix^i.$$ So that $$a_i=\binom{10}{i}3^i5^{10-i}.$$ Then $$f(i):=a_i/a_{i+1}=\frac{5}{3}\cdot \frac{i+1}{10-i}.$$ This is an increasing function of $$i$$ (for $$0\leq i<10$$). The maximum of the $$a_i$$ is attained at the smallest $$i$$ for which $$f(i)>1$$. Solving $$f(i)=1$$ yields $$i=25/8$$, which is larger than $$3$$ but smaller than $$4$$. Therefore $$a_4$$ is the largest coefficient.

The coefficients have the form $$\dbinom{10}k5^k3^{10-k}$$ The ratio of two consecutive coefficients is: $$\frac{\dbinom{10}k5^k3^{10-k}}{\dbinom{10}{k+1}5^{k+1}3^{9-k}}=\frac{3(k+1)}{5(10-k)}$$ This ratio is lesser than $$1$$ when $$3k+3<50-5k$$ That is, when $$k\le 5$$. This means that the maximum coefficient is $$\dbinom{10}65^6\cdot3^4$$.

• According to OP question, it would be more logical to define $k$ as the "power of $x$": $a_k x^k$ – Damien Jan 10 at 15:14

With the result proved below, the largest coefficient is among the set $$(a^n,b^n,C_p)$$, where $$C_p$$ is the largest coefficient as $$p\in[1,n-1]$$.

In this case, $$a=5, b=3, n=10$$, $$\displaystyle\frac ab>0$$. So by $$(1)$$,

$$\frac{an-b}{a+b} $$C_p={10\choose6}5^63^4>5^{10}>3^{10}$$

So the largest coefficient occurs at $$k=6$$.

Proof

By Binomial Formula,

$$(ax+by)^n=\sum_{k=0}^n {n\choose k}(ax)^k (by)^{n-k}$$

The coefficients can be expressed as

$$C_k={n\choose k}a^k b^{n-k}$$

Consider for all $$k, one may prove that

$$\frac{C_{k+1}}{C_k}=\frac ab\cdot\frac{n-k}{k+1}$$

This ratio, $$\widetilde C$$, indicates if $$|C_{k+1}|>|C_k|$$. This is true if $$|\widetilde C|>1$$.

Now if there exists an integer $$p\in[1,n-1]$$ such that $$C_{p}>0$$ is maximum, then

$$\left|\frac{C_{p}}{C_{p-1}}\right|>1, \left|\frac{C_{p+1}}{C_{p}}\right|<1$$

For $$\displaystyle\frac ab>0$$,

$$\frac ab n-1<(1+\frac ab)p<\frac ab(n+1)$$

$$\implies\frac{an-b}{a+b}

For $$\displaystyle-1<\frac ab<0$$,

$$\frac ab(n+1)<(1+\frac ab)p<\frac ab n-1$$

$$\implies\frac{an+a}{a+b}

For $$\displaystyle\frac ab<-1$$,

$$\frac ab n-1>(1+\frac ab)p; (\frac ab-1)p<\frac ab(n+1)$$

$$\implies\frac{an-b}{a+b}

The final step is to choose the largest coefficient among the set

$$(C_0=b^n, C_p, C_n=a^n)\tag4$$