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1967 HSC help (1 Viewer)
- Thread starter mathpie
- Start date
dan964
what
i will check if I have anything.
pretty sure though that everything I have is up.
pretty sure though that everything I have is up.
fluffchuck
Active Member
Consider:
m.a^m = a^m + a^m + ... + a^m (m terms)
m.a^m > a^0 + a^1 + ... + a^(m-1) (since a>0 and m is a positive integer)
m.a^m + [m.a^(m-1) + ... + m] > a^0 + a^1 + ... + a^(m-1) + [m.a^(m-1) + ... + m]
m.a^m + m.a^(m-1) + ... + m > 1 + a^1 + ... + a^(m-1) + m[a^(m-1) + ... + 1]
m.a^m + m.a^(m-1) + ... + m > 1 + a^1 + ... + a^(m-1) + m[a^(m-1) + ... + 1]
m(a^m + a^(m-1) + ... + 1) > (m+1)(1 + a^1 + ... + a^(m-1))
(a^m + a^(m-1) + ... + 1)/(m+1) > (1 + a^1 + ... + a^(m-1))/(m)
m.a^m = a^m + a^m + ... + a^m (m terms)
m.a^m > a^0 + a^1 + ... + a^(m-1) (since a>0 and m is a positive integer)
m.a^m + [m.a^(m-1) + ... + m] > a^0 + a^1 + ... + a^(m-1) + [m.a^(m-1) + ... + m]
m.a^m + m.a^(m-1) + ... + m > 1 + a^1 + ... + a^(m-1) + m[a^(m-1) + ... + 1]
m.a^m + m.a^(m-1) + ... + m > 1 + a^1 + ... + a^(m-1) + m[a^(m-1) + ... + 1]
m(a^m + a^(m-1) + ... + 1) > (m+1)(1 + a^1 + ... + a^(m-1))
(a^m + a^(m-1) + ... + 1)/(m+1) > (1 + a^1 + ... + a^(m-1))/(m)
The left hand side is the average of (m+1) numbers {a^0,a^1,a^2,...,a^(m)} and the right hand side is the average of m numbers {a^0,...,a^(m-1)}. Since a^m is bigger than all the other numbers, having that present will increase the average. You can instead prove that adding a number greater than the average to any set of (n) numbers will increase the average of the (n+1) numbers.