Terms for Reaching 10^57 in a Series?

[kurious]

How many terms would I need in the sum:
3 + 12 + 27 + 48 + ...

( 1x + 4x + 9x +16x + ...)

to get close to the number 10^57 ?

 

[Hurkyl]

Use the sum of squares formula.

1^2 + 2^2 + ... + n^2 = n (n+1) (2n+1) / 6

 

[arildno]

How many terms would I need in the sum:
3 + 12 + 27 + 48 + ...

( 1x + 4x + 9x +16x + ...)

to get close to the number 10^57 ?

Actually, if you allow an error margin of 10^58, one term is enough!
Since most numbers are larger than 10^58, 3 is a pretty close approximation to 10^57..

 

[kurious]

Thanks for helping.