$$\left(\dfrac{3}{x^2}-4x^3\right)^{10}$$
$$\mbox{The (r+1)th term will be }\quad 0\le{r}\le10\quad r\in \mathBb{Z}$$
$$10Cr\times \left(\frac{3}{x^2}\right)^r\times (-4x^3)^{10-r}\\
=10Cr \times 3^r\times(-4)^{10-r}\times \dfrac{1}{x^{2r}}\times (x^3)^{10-r}\\
=10Cr \times 3^r\times(-4)^{10-r}\times x^{-2r}\times x^{30-3r}\\
=10Cr \times 3^r\times(-4)^{10-r}\times x^{-2r+30-3r}\\
=10Cr \times 3^r\times(-4)^{10-r}\times x^{30-5r}$$
You want the constant term so 30-5r=0, r=6
so the term we want is
$$10C6\times 3^6\times (-4)^4 \times x^0$$
etc (correct assuming I didn't make any stupid errors)
$$\left(\dfrac{3}{x^2}-4x^3\right)^{10}$$
$$\mbox{The (r+1)th term will be }\quad 0\le{r}\le10\quad r\in \mathBb{Z}$$
$$10Cr\times \left(\frac{3}{x^2}\right)^r\times (-4x^3)^{10-r}\\
=10Cr \times 3^r\times(-4)^{10-r}\times \dfrac{1}{x^{2r}}\times (x^3)^{10-r}\\
=10Cr \times 3^r\times(-4)^{10-r}\times x^{-2r}\times x^{30-3r}\\
=10Cr \times 3^r\times(-4)^{10-r}\times x^{-2r+30-3r}\\
=10Cr \times 3^r\times(-4)^{10-r}\times x^{30-5r}$$
You want the constant term so 30-5r=0, r=6
so the term we want is
$$10C6\times 3^6\times (-4)^4 \times x^0$$
etc (correct assuming I didn't make any stupid errors)
Math(Input=Result) Error!
$${binom}{\left({\left({\frac{{\mathtt{3}}}{{{\mathtt{x}}}^{{\mathtt{2}}}}}{\mathtt{\,-\,}}{\mathtt{4}}{\mathtt{\,\times\,}}{{\mathtt{x}}}^{{\mathtt{3}}}\right)}^{{\mathtt{10}}}\right)} = {binom}{\left(\tiny\text{Error: loop}\right)}$$
$$\left(\dfrac{3}{x^2}-4x^3\right)^{10}\\$$
I think the answer is:
$$10Cr\times \left(\dfrac{3}{x^2}\right)^{\textcolor[rgb]{1,0,0}{10-r}}\times (-4x^3)^{\textcolor[rgb]{1,0,0}{r}}\\
=10Cr\times 3^{10-r} \times (-4)^r \times x^{-2(10-r)} \times x^{3r}\\
=10Cr\times 3^{10-r} \times (-4)^r \times x^{-2(10-r)+3r}\\
=10Cr\times 3^{10-r} \times (-4)^r \times x^{-20+2r+3r}\\
=10Cr\times 3^{10-r} \times (-4)^r \times x^{-20+5r}$$
so -20+5r=0 -> r = 4
the term we want is
$$10C\textcolor[rgb]{1,0,0}{4}\times 3^6\times (-4)^4 \times x^0$$
Using sigma notation and factorials for the combinatorial numbers, here is the binomial theorem:
Hi Heureka,
Mine is not wrong.
nCr is always identical to in value to nC(n-r)
10C4=10C6
your answer and my answer are identical.
Hi Melody,
your answer is not wrong!
But r ist also a position in the Pascal triangle too.
$$\\\mbox{r = 0: } (\frac{3}{x^2})^{10}=\dfrac{59049}{x^{20}}\\
\mbox{r = 1: } \dfrac{ -787320*x^{5} }{x^{20}}\\
\mbox{r = 2: } \dfrac{ 4723920*x^{10} }{x^{20}}\\
\mbox{r = 3: } \dfrac{ -16796160*x^{15} }{x^{20}}\\
\textcolor[rgb]{1,0,0}{\mbox{r = 4: } \dfrac{ 39191040*x^{20} }{x^{20}} = 39191040} \\
\mbox{r = 5: } \dfrac{ -62705664*x^{25} }{x^{20}}\\
\mbox{r = 6: } \dfrac{ 69672960*x^{30} }{x^{20}}\\
\mbox{r = 7: } \dfrac{ -53084160*x^{35} }{x^{20}}\\
\mbox{r = 8: } \dfrac{ 26542080*x^{40} }{x^{20}}\\
\mbox{r = 9: } \dfrac{ -7864320*x^{45} }{x^{20}}\\
\mbox{r = 10: } (-4x^3)^{10}=\dfrac{ 1048576*x^{50} }{x^{20}}$$
Many Greetings
Heureka