F4 Summation questions

21d)

(1+x)^m = 1 + a1 x + a2 x^2 + a3 x^3 + ... ar x^r + ... + x^m
(1+x)^n = 1 + b1 x + b2 x^2 + a3 x^3 + ... br x^r + ... + x^n
(1+x)^(m+n) = 1 + c1 x + c2 x^2 + c3 x^3 + ... cr x^r + ... + x^(m+n)


After expanding (1+x)^m (1+x)^n and comparing the coefficients.


we get
c1 = a1 + b1
c2 = a2 + a1b1 + b2
c3 = a3 + a2b1 + a1b2 + b3
c4 = a4 + a3b1 + a2b2 + a1b3 + b4
...
cr = ar + ar-1b1 + ar-2b2 + ar-3b3 + ... + a1br-1 + br


Then we can write it into summation form:


cr = summation{ ar-k bk }(where k = 0 to r) and a0 = b0 = 1


21e)


Since,
cr = m+n C r
ar = m C r
br = n C r


i.e.


m+n Cr = summation{ mCr-k nCk}


22a)
summation{ [k^3 - (k-1)^3] }(k = 1 to n)

= summation{[k^3]} - sumation{[(k-1)^3]}

= (1^3 + 2^3 + 3^3 + 4^3 + ... + n^3 ) - (0^3 + 1^3 + 2^3 + 3^3 + ... + (n-1)^3)

= 1^3 - 1^3 + 2^3 - 2^3 + 3^3 - 3^3 + ... + (n-1)^3 - (n-1)^3 + n^3

= n^3