The general expansion of a binomial raised to a power is:
(a+b)n=an+nan-1b+n(n-1)an-2b2/2!+n(n-1)(n-2)an-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=17¼, 4√15=15¼and 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 16¼=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)an-1b+C(n,2)an-2b2/2!+C(n,3)an-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).