$\begin{aligned}\left(2x+3y\right)^{2} & =\left(2x\right)^{2}+2\cdot(2x)(3y)+\left(3y\right)^{2}\\
& =4x^{2}+12xy+9y^{2}
\end{aligned}
$

$\begin{aligned}\left(a+b\right)^{n} & =\underset{k=0}{\overset{n}{\sum}}C_{k}^{n}\cdot a^{n-k}b^{k}\\
\left(a+b\right)^{3} & =C_{0}^{3}\cdot a^{3}+C_{1}^{3}\cdot a^{2}b+C_{2}^{3}\cdot ab^{2}+C_{3}^{3}\cdot b^{3}\\
& =a^{3}+3a^{2}b+3ab^{2}+b^{3}
\end{aligned}
$

$\left(a+b\right)^{n}=\underset{k=0}{\overset{n}{\sum}}C_{k}^{n}.a^{n-k}b^{k}$

$\left(a-b\right)^{5}=\left(a+(-b)\right)^{5}$

$=C_{0}^{5}a^{5}+C_{1}^{5}a^{4}(-b)$$+C_{2}^{5}a^{3}(-b)^{2}+C_{3}^{5}a^{2}(-b)^{3}$$+C_{4}^{5}a(-b)^{4}+C_{5}^{5}(-b)^{5}$

$\left(a-b\right)^{5}=\frac{5!}{5!\cdot0!}\cdot a^{5}+\frac{5!}{4!\cdot1!}\cdot-a^{4}b$$+\frac{5!}{3!\cdot2!}\cdot a^{3}b^{2}$$+\frac{5!}{3!\cdot2!}\cdot-a^{2}b^{3}+\frac{5!}{4!}\cdot ab^{4}$$+\frac{5!}{5!\cdot0!}\cdot-b^{5}$

$\left(a-b\right)^{5}=a^{5}-5a^{4}b$$+10a^{3}b^{2}-10a^{2}b^{3}$$+5ab^{4}-b^{5}.$

$\left(a+b\right)^{n}=\underset{k=0}{\overset{n}{\sum}}C_{k}^{n}\cdot a^{n-k}b^{k}$

$\left(2x-y\right)^{3}=\left(2x+(-y)\right)^{3}$

$=C_{0}^{3}(2x)^{3}+C_{1}^{3}(2x)^{2}(-y)$$+C_{2}^{3}(2x)(-y)^{2}+C_{3}^{3}(-y)^{3}$

$\left(2x-y\right)^{3}=\frac{3!}{3!\cdot0!}\cdot8x^{3}+\frac{3!}{2!\cdot1!}\cdot4x^{2}(-y)$$+\frac{3!}{2!\cdot1!}\cdot2xy^{2}+\frac{3!}{3!\cdot0!}\cdot(-y)^{3}$

$=8x^{3}-12x^{2}y+6xy^{2}-y^{3}$

$\left(a+b\right)^{n}=\underset{k=0}{\overset{n}{\sum}}C_{k}^{n}\cdot a^{n-k}b^{k}$

$\left(2x-5y\right)^{4}=C_{0}^{4}\cdot\left(2x\right)^{4}+C_{1}^{4}\cdot(2x)^{3}\cdot\left(-5y\right)$$+C_{2}^{4}\cdot(2x)^{2}(-5y)^{2}$$+C_{3}^{4}\cdot(2x)(-5y)^{3}+C_{4}^{4}\cdot(-5y)^{4}$

$\left(2x-5y\right)^{4}=1\cdot16x^{2}+4\cdot8x^{3}\cdot(-5y)$$+6\cdot4x^{2}\cdot25y^{2}$$+4\cdot2x\cdot(-125y^{3})+1\cdot625y^{4}$

$\left(2x-5y\right)^{4}=16x^{2}-160x^{3}y$$+600x^{2}y^{2}-1000xy^{3}$$+625y^{4}.$

$\left(a+b\right)^{n}=\underset{k=0}{\overset{n}{\sum}}C_{k}^{n}\cdot a^{n-k}b^{k}$

Cara mencari suku ke-p adalah : $C_{p-1}^{n}\cdot a^{n-(p-1)}b^{p-1}$

Suku ke-lima dari penjabaran $\left(3x+y\right)^{7}$ :

$\begin{aligned}C_{4}^{7}\cdot\left(3x\right)^{7-4}y^{4} & =\frac{7!}{(7-4)!\cdot4!}\cdot\left(3x\right)^{3}y^{4}\\
& =\frac{7\cdot6\cdot5\cdot4!}{4!\cdot3\cdot2\cdot1}\cdot27x^{3}\cdot y^{4}\\
& =945x^{3}y^{4}.
\end{aligned}
$

Jadi suku kelima nya adalah $945x^{3}y^{4}.$

$\left(a+b\right)^{n}=\underset{k=0}{\overset{n}{\sum}}C_{k}^{n}\cdot a^{n-k}b^{k}$

Cara mencari suku ke-p adalah : $C_{p-1}^{n}\cdot a^{n-(p-1)}b^{p-1}$

Suku ke-enam dari $\left(\frac{1}{a}-a\right)^{10}:$

$\begin{aligned}C_{5}^{10}\left(\frac{1}{a}\right)^{10-5}(-a)^{5} & =\frac{10!}{(10-5)!\cdot5!}\cdot\frac{1}{a^{5}}\cdot(-a^{5})\\
& =\frac{10\cdot9\cdot8\cdot7\cdot6\cdot5!}{5!\cdot5\cdot4\cdot3\cdot2\cdot1}\cdot(-1)\\
& =-252
\end{aligned}
$

Jadi suku keenam dari $\left(\frac{1}{a}-a\right)^{10}$ adalah $-252.$

$\left(a+b+c\right)^{3}=\left\{ a+(b+c)\right\} ^{3}$

$\left\{ a+(b+c)\right\} ^{3}=a^{3}+3a^{2}\left(b+c\right)$$+3a(b+c)^{2}+(b+c)^{3}$

$\left\{ a+(b+c)\right\} ^{3}=a^{3}+3a^{2}b+3a^{2}c$$+3a(b^{2}+2bc+c^{2})$$+(b^{3}+3b^{2}c+3bc^{2}+c^{3})$

$\left\{ a+(b+c)\right\} ^{3}=a^{3}+b^{3}+c^{3}$$+3a^{2}b+3a^{2}c$$+3ab^{2}+6abc$$+3ac^{2}+3b^{2}c+3bc^{2}.$

$\left(a+b\right)^{n}=\underset{k=0}{\overset{n}{\sum}}C_{k}^{n}a^{n-k}b^{k}$

Cara mencari suku ke-p adalah : $C_{p-1}^{n}\cdot a^{n-(p-1)}b^{p-1}$

Suku ke-tujuh dari $\left(x^{3}-\frac{1}{x}\right)^{8}$:

$\begin{aligned}C_{6}^{8}\cdot\left(x^{3}\right)^{8-6}\cdot\left(-\frac{1}{x}\right)^{6} & =\frac{8!}{6!\cdot2!}.x^{6}\cdot\left(\frac{1}{x^{6}}\right)\\
& =\frac{8\cdot7\cdot6!}{6!\cdot2\cdot1}\cdot1\\
& =28
\end{aligned}
$

Jadi suku ketujuhnya adalah $28.$

Karena berpangkat enam, maka banyaknya suku ada $7$. Dengan demikian suku tengahnya adalah suku ke-empat

$\left(a+b\right)^{n}=\underset{k=0}{\overset{n}{\sum}}C_{k}^{n}\cdot a^{n-k}b^{k}$

Cara mencari suku ke-p adalah : $C_{p-1}^{n}\cdot a^{n-(p-1)}b^{p-1}$

Suku ke-empat dari $\left(2x^{3}-\sqrt{x}\right)^{6}$:

$\begin{aligned}C_{3}^{6}\cdot\left(2x^{3}\right)^{6-3}\cdot\left(-\sqrt{x}\right)^{3} & =\frac{6!}{3!\cdot3!}\cdot\left(2x^{3}\right)^{3}\cdot(-x\sqrt{x})\\
& =\frac{6\cdot5\cdot4\cdot3!}{3!\cdot3\cdot2\cdot1}\cdot8x^{9}\cdot(-x\sqrt{x})\\
& =-160x^{10}\sqrt{x}
\end{aligned}
$

Jadi koefisien dari suku tengah ekspansi binom $\left(2x^{3}-\sqrt{x}\right)^{6}$ adalah $-160.$

$\left(a+b\right)^{n}=\underset{k=0}{\overset{n}{\sum}}C_{k}^{n}a^{n-k}b^{k}$

Cara mencari suku ke-p adalah : $C_{p-1}^{n}\cdot a^{n-(p-1)}b^{p-1}$

$x^{3}y^{2}$berada pada suku ketiga $\left(2x+3y\right)^{5}$ sehingga :

$\begin{aligned}C_{2}^{5}\cdot(2x)^{5-2}\cdot(3y)^{2} & =\frac{5!}{(5-2)!\cdot2!}\cdot(2x)^{3}\cdot(3y)^{2}\\
& =\frac{5\cdot4\cdot3!}{3!\cdot2!}\cdot8x^{3}\cdot9y^{2}\\
& =10\cdot8x^{3}\cdot9y^{2}\\
& =720x^{3}y^{2}
\end{aligned}
$

Jadi koefisien dari $x^{3}y^{2}$ adalah $720.$

$\begin{aligned}\left(1-2x\right)^{10} & =\underset{k=0}{\overset{10}{\sum}}C_{k}^{10}1^{10-k}(-2x)^{k}\\
& =\underset{k=0}{\overset{n}{\sum}}C_{k}^{10}\cdot(-2)^{k}.x^{k}
\end{aligned}
$

Suku yang mengandung $x^{6}$ diperoleh jika $x^{6}=x^{k},$ sehingga diperoleh $k=6$

Koefisien $x^{6}$yaitu :

$\begin{aligned}C_{6}^{10}(-2)^{6} & =\frac{10!}{6!\cdot4!}\cdot2^{6}\\
& =\frac{10\cdot9\cdot8\cdot5\cdot7\cdot6!}{6!\cdot4\cdot3\cdot2\cdot1}\cdot2^{6}\\
& =210\cdot64\\
& =13.440
\end{aligned}
$

Jadi koefisien dari $x^{6}$ adalah $13.440.$

$\left(x^{2}-2y\right)^{6}$= $\underset{k=0}{\overset{6}{\sum}}C_{k}^{6}\cdot(x^{2})^{6-k}\cdot(-2y)^{k}$

Suku yang mengandung $x^{6}$

$(x^{2})^{6-k}=x^{6}$

$2(6-k)=6$

$6-k=3$

$k=3$

Koefisien yang mengandung $x^{6}$adalahh :

$\begin{aligned}C_{k}^{6}\cdot(x^{2})^{6-k}\cdot(-2y)^{k} & =C_{3}^{6}\cdot(x^{2})^{6-3}\cdot(-2y)^{3}\\
& =\frac{6!}{3!\cdot3!}\cdot x^{6}\cdot(-8y^{3})\\
& =-160x^{6}y^{3}
\end{aligned}
$

Jadi koefisien $x^{6}$adalah $-160.$

$\begin{aligned}\left(2x^{2}-y^{3}\right)^{8} & =\underset{k=0}{\overset{8}{\sum}}C_{k}^{8}(2x^{2})^{8-k}\cdot(-y^{3})^{k}\\
& =\underset{k=0}{\overset{n}{\sum}}2^{8-k}\cdot\left(x^{2}\right)^{8-k}\cdot(-y^{3})^{r}
\end{aligned}
$

Suku yang mengandung $x^{10}$ didapat jika :

$\begin{aligned}\left(x^{2}\right)^{8-k} & =x^{10}\\
2(8-k) & =10\\
16-2k & =10\\
2k & =6\\
k & =3
\end{aligned}
$

Jadi suku pada penjabaran $\left(2x^{2}-y^{3}\right)^{8}$ yang mengandung $x^{10}$ adalah $C_{3}^{8}\cdot2^{8-3}.x^{16-2\cdot3}\left(-y^{3}\right)^{3}=-1.792x^{10}y^{9}.$

$\begin{aligned}\left(5+2x\right)^{n} & =\underset{k=0}{\overset{n}{\sum}}C_{k}^{n}\cdot5^{n-r}\cdot\left(2x\right)^{r}\\
& =\underset{k=0}{\overset{n}{\sum}}5^{n}\cdot5^{-r}\cdot2^{r}\cdot x^{r}
\end{aligned}
$

Koefisien pada suku ke-$x^{2}$adalah : $C_{2}^{n}\cdot5^{n}\cdot5^{-2}\cdot2^{2}$

Koefisen pada suku $x$ adalah : $C_{1}^{n}\cdot5^{n}\cdot5^{-1}\cdot2$

koefisien pada suku $x^{2}=$ koefisien pada suku $x$

$C_{2}^{n}\cdot5^{n}\cdot5^{-2}\cdot2^{2}=C_{1}^{n}\cdot5^{n}\cdot5^{-1}\cdot2$

$\frac{n!}{(n-2)!.2!}\cdot\frac{2}{5}=\frac{n!}{(n-1)!}$

$\frac{n(n-1)(n-2)!}{(n-2)!\cdot2\cdot1}\cdot\frac{2}{5}=\frac{n(n-1)!}{(n-1)!}$

$\frac{n(n-1)}{5}=n$

$n^{2}-n=5n$

$n^{2}-6n=0$

$n\left(n-6\right)=0$

$n=0$ (tidak memenuhi)

$n=6$ (memenuhi)

Jadi nilai $n$ yang memenuhi adalah $6.$