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# Algebra

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Find the coefficient of y^4 in the expansion of (2y-7)^5.

Jun 14, 2022

#1
-1

$It's \ literally \ 1 \ dumbbo$

Jun 14, 2022
#2
+556
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I'm sorry to say this but, think again guest answerer

$$\mathrm{Apply\:binomial\:theorem}:\quad \left(a+b\right)^n=\sum _{i=0}^n\binom{n}{i}a^{\left(n-i\right)}b^i$$

(*ahhhh man not this again...)

Just know that:

$$a=2y,\:\:b=-7$$

when we are substituting values in the formula!!!

$$=\sum _{i=0}^5\binom{5}{i}\left(2y\right)^{\left(5-i\right)}\left(-7\right)^i$$

$$=\frac{5!}{0!\left(5-0\right)!}\left(2y\right)^5\left(-7\right)^0+\frac{5!}{1!\left(5-1\right)!}\left(2y\right)^4\left(-7\right)^1+\frac{5!}{2!\left(5-2\right)!}\left(2y\right)^3\left(-7\right)^2+\frac{5!}{3!\left(5-3\right)!}\left(2y\right)^2\left(-7\right)^3+\frac{5!}{4!\left(5-4\right)!}\left(2y\right)^1\left(-7\right)^4+\frac{5!}{5!\left(5-5\right)!}\left(2y\right)^0\left(-7\right)^5$$

$$32y^5-560y^4+3920y^3-13720y^2+24010y-16807$$

The coefficient of y^4 in the expansion of (2y-7)^5 is $$-560$$

-Vinculum

Jun 14, 2022
#3
-1

Guest Jun 14, 2022
#4
+556
+1

If you think you are right

-Vinculum

Vinculum  Jun 14, 2022
#5
+280
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hipie  Jun 14, 2022
edited by Guest  Jun 14, 2022
#6
0

I think there was something wrong with the binomial expansion

Guest Jun 14, 2022
#7
+556
+1

Sorry Bro lemme check again

Vinculum  Jun 14, 2022
edited by Vinculum  Jun 14, 2022
#8
+556
+1

Edit: I double checked in Wolfram|Alpha.

It should be right.

https://www.wolframalpha.com/input?i2d=true&i=Power%5B%5Cleft%5C%2840%292y-7%5Cright%5C%2841%29%2C5%5D

Jun 14, 2022
#9
0

$the \ answer \ is \ incorrect, \ face \ it.$

Guest Jun 14, 2022
#10
+556
+1

Well, I did everything I can, I don't think I can solve this one then. Maybe ask someone or a mod !

-Vinculum

Jun 14, 2022