BINOMIAL THEOREM

18) in the binomial expansion of (1 + x/n)^n in ascending powers of x,the coefficient of x^2 is 3/7.
(a) find the value of n
(b) find the coefficient of x^3.
(a)
(1 + x/n)^n
=1+x+nC2(x/n)^2+nC3(x/n)^3
=1+x+[(n-1)/2n]x^2+[(n-1)(n-2)/6n^2]x^3
the coefficient of x^2 is 3/7.
(n-1)/2n=3/7
7n-7=6n
n=7
(b)
the coefficient of x^3
=[(n-1)(n-2)/6n^2]
=(6*5)/(6*49)
=5/49
19) (a) expand (a+b)^5 + (a-b)^5
(b) hence,evaluate (開方3 +1)^5 + (開方3 - 1)^5.leave your answer in surd form.
(c) hence,form a quadratic equation whose roots are (開方3 +1)^5 and (開方3 - 1)^5
(a)
(a+b)^5 + (a-b)^5
=(a^5+5a^4b+10a^3b^2+10a^2b^3+5ab^4+b^5)+(a^5-5a^4b+10a^3b^2-10a^2b^3+5ab^4-b^5)
=2a^5+20a^3b+10ab^4
(b)
(開方3 +1)^5 + (開方3 - 1)^5
a=√3, b=1
(√3 +1)^5 + (√3 - 1)^5
=2(√3)^5+20(√3)^3+10(√3)
=18√3+60√3+10√3
=88√3
(c)
(√3 +1)^5 (√3 - 1)^5
=[(√3 +1)(√3 - 1)]^5
=(3-1)^5
=2^5
=32
So a quadratic equation whose roots are (開方3 +1)^5 and (開方3 - 1)^5 is
x^2-88√3x+32