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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?

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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.

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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 $\dbinom{10}65^6\cdot3^4$.

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  • $\begingroup$ According to OP question, it would be more logical to define $k$ as the "power of $x$": $a_k x^k$ $\endgroup$ – Damien Jan 10 at 15:14
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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 $(1)$,

$$\frac{an-b}{a+b}<p<\frac{an+a}{a+b}\implies\frac{47}{8}<p<\frac{55}{8}\implies p=6$$ $$C_p={10\choose6}5^63^4>5^{10}>3^{10}$$

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 $k<n$, 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 $\displaystyle\frac ab>0$,

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

$$\implies\frac{an-b}{a+b}<p<\frac{an+a}{a+b}\tag1$$

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

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

$$\implies\frac{an+a}{a+b}<p<\frac{an-b}{a+b}\tag2$$

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

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

$$\implies\frac{an-b}{a+b}<p<\frac{an+a}{a-b}\tag3$$

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

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

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