F.4 AM

1
Let the number is x,y
Then x^2+y^2=(x-y)^2+2xy
Since (x-y)^2>=o, when x=y we get the minimum value of x^2+y^2
From x+y=10, we have x=y=5
When the two numbers are 5 , 5 the sum of their squares is a minimum.
2
(a)
First term p
Second term (p-2)p/1!
Third term (p-4)p(p -1)/2!
We deduce that
k th term (p-2(k-1))p(p-1)(p-2)...(p-(k-2))/(k-1)!
k+1 th term (p-2k)p(p-1)(p-2)...(p-(k-1))/k!
(b)
Let P(n) be the statement
p+(p-2)p/1!+(p-4)p(p -1)/2!+(p-6)p(p-1)(p -2)/3!+.....(p-2(n-1))p(p-1)(p-2)...(p-(n-2))/(n-1)!=p(p-1)(p-2)....(p-(n- 1))/(n-1)!
when n=1
LHS=p
RHS=p
when n=1 P(1) is true
when n=k assume that P(k) is true
when n=k+1
LHS
=p+(p-2)p/1!+(p-4)p(p -1)/2!+(p-6)p(p-1)(p -2)/3!+.....(p-2(k-1))p(p-1)(p-2)...(p-(k-2))/(k-1)!+ (p-2k)p(p-1)(p-2)...(p-(k-1))/k!
=p(p-1)(p-2)....(p-(k- 1))/(k-1)!+ (p-2k)p(p-1)(p-2)...(p-(k-1))/k!
=p(p-1)(p-2)....(p-(k- 1))[1/(k-1)!+ (p-2k)/k!]
=p(p-1)(p-2)....(p-(k- 1))[(k+p-2k)/k!]
=p(p-1)(p-2)....(p-(k- 1))(p-k)/k!
=RHS
when n=k, P(k) is true
By MI for all positive value of n, P(n) is true