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

 Jun 14, 2022
 #1
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-1

$It's \ literally \ 1 \ dumbbo$

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

 

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

 

(*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
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-1

the answer was wrong

Guest Jun 14, 2022
 #4
avatar+556 
+1

If you think you are right

 

Prove it, guest answerer...

 

cool

 

-Vinculum

Vinculum  Jun 14, 2022
 #5
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No I was the asker and your answer was incorrect, chill.

hipie  Jun 14, 2022
edited by Guest  Jun 14, 2022
 #6
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I think there was something wrong with the binomial expansion

Guest Jun 14, 2022
 #7
avatar+556 
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Sorry Bro lemme check again

Vinculum  Jun 14, 2022
edited by Vinculum  Jun 14, 2022
 #8
avatar+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
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0

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

Guest Jun 14, 2022
 #10
avatar+556 
+1

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

 

 

-Vinculum

 Jun 14, 2022

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