The sum of the squares of the first ten natural numbers is,
$$1^2 + 2^2 + ... + 10^2 = 385$$The square of the sum of the first ten natural numbers is,
$$(1 + 2 + ... + 10)^2 = 55^2 = 3025$$Hence the difference between the sum of the squares of the first ten natural numbers and the square of the sum is $$3025 − 385 = 2640$$.
Find the difference between the sum of the squares of the first one hundred natural numbers and the square of the sum.
For summing the first n numbers the formula is simple: $$ sum(1+2+...+n)=\frac{n*(n+1)}{2}$$
In [2]:
def sum_first_n(n):
return n*(n+1)/2
now the square of the items is again provable by induction that it is: $$sum(1^2+2^2+..+n^2)= \frac{n(n+1)(2n+1)}{6}$$
In [4]:
def sum_square_first_n(n):
return n*(n+1)*(2*n+1)/6
so the answer is:
In [7]:
n = 100
print(int(pow(sum_first_n(n),2) - sum_square_first_n(n)))