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Doing this one question for an hour now. See no hope. While im at it, can someone see if they can solve this?

 

The first three terms in the expansion of (1+kX)^n are 1 -4X + (15/2)X^2. Find k and n.

First term: 1

Second term: -4X

Third term: (15/2)X^2

 

Keep in mind the pascal triangle and binomial coefficient. I tried using graph but no hope. I don't want to sink low putting in random numbers and pray.

DeadRight  Jul 17, 2017

Best Answer 

 #1
avatar+25874 
+1

We have \((1+kX)^n\rightarrow 1+nkX+\frac{n(n-1)}{2}(kX)^2+...\)

 

So \(nk=-4\\\frac{n(n-1)}{2}k^2=\frac{15}{2} \)

 

You now have two equations for two unknowns. Can you take it from here?

Alan  Jul 17, 2017
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1+0 Answers

 #1
avatar+25874 
+1
Best Answer

We have \((1+kX)^n\rightarrow 1+nkX+\frac{n(n-1)}{2}(kX)^2+...\)

 

So \(nk=-4\\\frac{n(n-1)}{2}k^2=\frac{15}{2} \)

 

You now have two equations for two unknowns. Can you take it from here?

Alan  Jul 17, 2017

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