The ﬁrst three terms in the expansion of(2+ax)
n, in ascending powers of x, are 32−40x+bx2. Find the values of the constants n, a and b.

Question

The ﬁrst three terms in the expansion of(2+ax)
n, in ascending powers of x, are 32−40x+bx2. Find the values of the constants n, a and b.

anonymous · Answer

expanding (2+ax)^n we have 2^n + n*a*x +n*(n-1)/2 *a^2x^2 .....
thus we have 2^n =32 but we know 32 =2^5
hence n=5

anonymous · Answer

again n*a=-40 as we have n=5  hence we have a=-8

anonymous · Answer

thus b=n*(n-1)/2 *a^2 =5*4/2 *(-8)^2=640

anonymous · Answer

@matricked I get the first step for the second binomial term isn't it n*2^(n-1)ax=-40x?

anonymous · Answer

The answers say a = -1/2 and b = 20 . I understand how a = -1/2 but I don't for b

campbell_st · Answer

well n = 5 since 2^5 = 32

the  binomial expansion is

$$^5Cn(2)^{5 - n}(bx)^n$$

where n = 0, 1, 2, 3, 4, 5

anonymous · Answer

So to find the 3rd term do I do 5C2(2)^3(ax)^2?

campbell_st · Answer

oops its 

$$^5C_{1} (2)^4 (ax)^1 = -40x$$

solve for x

anonymous · Answer

Yes I understand that

campbell_st · Answer

which is 

$$5 \times 16 \times ax = -40x$$

anonymous · Answer

So a = -0.5?

campbell_st · Answer

so the 3rd term is 

$$^5C_{2} (2)^3 (-\frac{1}{2}x)^2 = bx^2$$

campbell_st · Answer

yep a is definitely -1/2 

so the last part is 

$$10 \times 8 \times (-\frac{1}{2})^2 \times x^2 = bx^2$$

anonymous · Answer

And I presume b = 20. Thanks so much for your help!

campbell_st · Answer

well yes it does show b = 20... 

glad to help. 

$$\frac{80}{4}x^2 = bx^2$$