Showing binomial expansion and expressing in terms of factorials

Question

I need help with this question please.

Answer 1

The general expansion of a binomial raised to a power is:

(a+b)ⁿ=aⁿ+na^n-1b+n(n-1)a^n-2b²/2!+n(n-1)(n-2)a^n-3b³/3!+...

If a=1 then all powers of a are 1.

(i) In (1+x)^¼, a=1, b=x, n=¼ so the expansion is:

1+¼x+(¼)(-¾)x²/2!+(¼)(-¾)(-5/4)x³/3!+...

while, in (1-x)¼ is expanded as:

1-¼x+(¼)(-¾)x²/2!-(¼)(-¾)(-5/4)x³/3!+...

When we add the two series together alternate terms cancel out and we're left with:

2(1+(¼)(-¾)x²/2!+(¼)(-¾)(-5/4)(-7/4)x⁴/4!+...).

We can ignore all powers of x beyond x²:

2(1-3x²/32)=2-3x²/16.

(ii) Using the above formula, since ⁴√17=17^¼, ⁴√15=15^¼and x=±1/16, 1+x=1¹/₁₆=17/16, 1-x=15/16:

(1+x)^¼+(1-x)^¼=(17/16)^¼+(15/16)^¼=2-3x²/16=1.99927 approx.

But (17/16)^¼+(15/16)^¼=⁴√17/2+⁴√15/2=(⁴√17+⁴√15)/2, and 16^¼=2.

To get ⁴√17+⁴√15 we need to multiply this result by 2.

Therefore ⁴√17+⁴√15=2×1.99927=3.9985 to 4 places of decimals. (The actual value is also 3.9985 to 4 decimal places.)

(c) This can sometimes be written C(n,r) and C(n,r-1). This notation convention makes it easier for me to show you and to explain the solution.

These are the coefficients in the binomial expansion. For example, the expansion of (a+b)ⁿ can be expressed as:

C(n,0)aⁿ+C(n,1)a^n-1b+C(n,2)a^n-2b²/2!+C(n,3)a^n-3b³/3!+...

If n=5, C(5,0)=1, C(5,1)=5, C(5,2)=5(5-1)/2!=10, C(5,3)=5(5-1)(5-2)/3!=10, C(5,4)=5(5-1)(5-2)(5-3)/4!=5, C(5,5)=1. I've used this example to show you how the combination function C (how many ways to combine r objects out of n objects) works and how it's related to the binomial expansion.

In general, C(n,r)=n!/(r!(n-r)!). So C(n,r-1) and C(n,r) are adjacent coefficients in the expansion.

C(n,r)+C(n,r-1)=n!/(r!(n-r)!)+n!/((r-1)!(n-r+1)!).

Now, r!=r[(r-1)!] and (n-r+1)!=(n-r+1)[(n-r)!].

We can rewrite C(n,r)+C(n,r-1)=

n!/r[(r-1)!](n-r)!)+n!/((r-1)!(n-r+1)[(n-r)!]), which factorises:

[n!/((r-1)!(n-r)!)][(1/r)+(1/(n-r+1))]=

[n!/((r-1)!(n-r)!)][(n-r+1+r)/(r(n-r+1))]=

[n!/((r-1)!(n-r)!)][(n+1)/(r(n-r+1))]=

(n+1)!/(r!(n-r+1)!).

C(n+1,r)=(n+1)!/(r!(n+1-r)! which is the same expression as above.

Therefore C(n,r)+C(n,r-1)=C(n+1,r).