|
I'm sorry I didn't get to this earlier but all those thirty second mystery posts have totally ruined this forum. It didn't look like there were any intersting threads in here.
Anyway, for your first problem, onionmon is correct and the number of terms you have will be n(n+1)/2. It just so happens that n is the number in the series you are looking for! So, the equation now becomes:
n(n+1)/2 = 1,000,000
...so isolate for n and you get...
n = (sqrt(1 + 8,000,000) - 1) / 2
...or 1413.71
Now, round this up because this formula assumes we get to sum the end of the n'th line and you discover that the millionth number is 1414...
Now, the second one is also not too hard. You just have to generalize the quadratic expansion. So, let me formulate some notation 'cause I can't just scribble into the TFP...
Let SC(a-z,2) be the sum of all possible products of 2 element subsets of the variables a through z. Furthermore, define SC(a-z,0) = 1, since the null set has no elements to create a sum of products.
For example, SC(a-z,2) would look like ab + ac + ... + az + bc + bd + ... + bz + ... + yz.
Let S(i,n)f(i) denote the sum f(0) + f(1) + ... + f(i) + ... + f(n).
This will be our sigma notation.
So, the product of (x + a)(x + b)...(x + z) is equal to:
S(i,26)( SC(a-z,i) * x^(26-i) )
Pretty simple, eh?
|