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Sagot :
The binomial theorem says that
[tex](x+y)^n=\sum ^n_{k\mathop=0}\frac{n!}{k!(n-k)!}x^{n-k}y^k[/tex]Looking at our problem we can see that
• x = 4t
,• y = -3
,• n = 5
Using it in the binomial theorem
[tex](4t-3)^5=\sum ^5_{k\mathop{=}0}\frac{5!}{k!(5-k)!}(4t)^{5-k}(-3)^k[/tex]And now we do the calculus for each term, then, I'll do it separated, just to be organized since binomial theorem calculus can be long, I'll start with k = 0 and go on.
k = 0
[tex]\begin{gathered} \frac{5!}{0!(5-0)!}(4t)^{5-0}(-3)^0=4^5t^5 \\ \\ 4^5t^5=1024t^5 \end{gathered}[/tex]k = 1
[tex]\begin{gathered} \frac{5!}{1!(5-1)!}(4t)^{5-1}(-3)^1=\frac{5!}{4!}4^4t^4(-3) \\ \\ \frac{5!}{4!}4^4t^4(-3)=(-3)\cdot5\cdot256\cdot t^4 \\ \\ (-3)\cdot5\cdot256\cdot t^4=-3840t^4 \end{gathered}[/tex]k = 2
[tex]\begin{gathered} \frac{5!}{2!(5-2)!}(4t)^{5-2}(-3)^2=\frac{5!}{2!\cdot3!}(4t)^3(-3)^2 \\ \\ \frac{5!}{2!\cdot3!}(4t)^3(-3)^2=10\cdot64\cdot t^3\cdot9 \\ \\ 10\cdot64\cdot t^3\cdot9=5760t^3 \end{gathered}[/tex]k = 3
[tex]\frac{5!}{3!(5-3)!}(4t)^{5-3}(-3)^3=-4320t^2[/tex]k = 4
[tex]\frac{5!}{4!(5-4)!}(4t)^{5-4}(-3)^4=1620t^{}[/tex]And the last one, k = 5
[tex]\frac{5!}{5!(5-5)!}(4t)^{5-5}(-3)^5=-243[/tex]Now we did all terms, the binomial will be the sum of all of them
[tex](4t-3)^5=1024t^5-3840t^4+5760t^3-4320t^2+1620t-243[/tex]
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