+0

# Im fatigued, someone help! Working with power.

0
334
1
+27

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.

Jul 17, 2017

#1
+27480
+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?

Jul 17, 2017

#1
+27480
+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