Binomial Theorem (1 Viewer)

[Kujah]

I'm stuck on a few questions, and I require some assistance

i. Obtain the term independent of x in (2-x+3x^2)(1-2/x)^9

ii. The expansion of (1+ax+bx^2)(1+x)^10 in ascending powers of x begins with the terms 1+x+x^2. Determine the values of a,b.

iii. The first three terms in the expansion of (1+cx)^n, where n is a postivie integer, are 1+6x+16x^2. Show that nc=6, n(n-1)c^2=32. Hence, find n and c.

Answers:

i. 452
ii. a=-9, b=46
iii. n=9, c=2/3

I've been trying for hours, but I've gotten nowhere

Cheers.

 

[ssglain]

i. Obtain the term independent of x in (2 - x + 3x^2)(1 - 2/x)^9

The term "independent of x" is one that looks like mx^0, where m is a real number.

Let's look at the terms in (2 - x + 3x²). We have 2, -x and 3x².
To obtain the term "independent of x", we need to multiply 2 by a constant term, -x by a term in 1/x and 3x² by a term in 1/x² and then add them together.

Let's consider the expansion of (1 - 2/x)^9
Any term can be written as 9Ck*[1^k][(-2)^(9 - k)][(1/x)^(9 - k)] = 9Ck*[(-2)^(9 - k)][x^(k - 9)]

x^(k - 9) = x^0 --> k - 9 = 0
.: k = 9
Hence 9C9*[(-2)^0](x^0) = 1

x^(k - 9) = x^-1 --> k - 9 = -1
.: k = 8
Hence 9C8*[(-2)^1](x^-1) = -18/x

x^(k - 9) = x^-2 --> k - 9 = -2
.: k = 7
Hence 9C7*[(-2)^2](x^-2) = 144/x²

.: The required term independent of x is 2(1) + (-x)(-18/x) + (3x²)(144/x²) = 2 + 18 + 432 = 452

 

[Forbidden.]

Learn your Binomial Theorems well because there didn't appear to be any in our paper but will be very likely to appear in your paper.

 

[ssglain]

ii. The expansion of (1 + ax + bx²)(1 + x)^10 in ascending powers of x begins with the terms 1 + x + x². Determine the values of a,b.

This one can be done by basically the same principle, but in addition to finding the term independent of x, we must now also find the terms in x and x². Then solving a couple of simulaneous equations should yield values for a and b. It is slightly more complicated, but try for yourself first. The solutions are behind the spoiler.

Spoiler

From (1 + ax + bx²), we have terms 1, ax and bx².
To obtain the term independent of x, we need to multiply 1 by a constant term.
To obtain the term in x, we need to multiply 1 by a term in x, ax by a constant term.
To obtain the term in x², we need to multiply 1 by a term in x², ax by a term in x, bx² by a constant term.

In the expansion of (1 + x)^10, any term can be written as 10Ck*(x^k).

For the constant term:
x^k = x^0
.: k = 0
i.e. 10C0*(x^0) = 1

For the term in x:
x^k = x^1
.: k = 1
i.e. 10C1*(x^1) = 10x

For the term in x²:
x^k = x²
.: k = 2
i.e. 10C2*(x²) = 45x²

It follows that in the expansion of (1 + ax + bx²)(1 + x)^10:
The term independent of x is (1*1) = 1
The term in x is (1*10x + ax*1) = (10 + a)x
The term in x² is (1*45x² + ax*10x + bx²*1) = (45 + 10a +b)x²

Equating the above with like terms in 1 + x + x²:
10 + a = 1 ... (1)
45 + 10a + b = 1 ... (2)

From (1), a = -9
Put in (2), 45 - 90 + b = 1 --> b = 46

iii. The first three terms in the expansion of (1 + cx)^n, where n is a postivie integer, are 1 + 6x + 16x². Show that nc=6, n(n - 1)c²=32. Hence, find n and c.

Spoiler

Any term in the expansion of (1 + cx)^n can be written as nCk*(c^k)(x^k).

For the term in x:
x^k = x^1
.: k = 1
i.e. 6x = nC1*cx
nC1*c = 6
[n!/(n - 1)!]c = 6
.: nc = 6 ... (1)

For the term in x²:
x^k = x^2
.: k = 2
i.e. 16x² = nC2*c²x²
nC2*c² = 16
n!/[2!(n - 2)!]c² = 16
.: n(n - 1)c² = 32 ... (2)

From (1), n²c² = 36 --> c² = 36/n² ... (3)

Put (3) in (2), 36(n - 1)/n = 32 --> 9n - 9 = 8n
.: n = 9

Put in (1), 9c = 6
.: c = 2/3

 

[Kujah]

Thanks a bunch! Have you got any advice for tackling these types of questions in Binomial Theorem, or MX1 in general?

 

[ssglain]

Thanks a bunch! Have you got any advice for tackling these types of questions in Binomial Theorem, or MX1 in general?

You're more than welcome.

Well, as you saw, those three questions were all solved by essentially the same principal - writing out the general term of the binomial expansion and using this to find specific terms.

I'm afraid that all I can offer on the forum are solutions and basic explanations to selected questions. I am providing private tuition in inner-west suburbs of Sydney, if you want more extensive help and advice for tackling this as well as other topics in MX1 and MX2. Please PM me or add me on MSN if you want to discuss this.