The sum of the first n odd numbers is n squared! So, the short answer is that the sum of the first 70 odd numbers is 70 squared, i.e. 4900.
Allow me to prove the result: odd numbers come in the form 2n-1, because 2n is always even, and the number immediately before an even number is always odd.
So, if we sum the first N odd numbers, we have
[tex]\displaystyle \sum_{i=1}^N 2i-1 = 2\sum_{i=1}^N i - \sum_{i=1}^N 1[/tex]
The first sum is the sum of all integers from 1 to N, which is N(N+1)/2. We want twice this sum, so we have
[tex]\displaystyle 2\sum_{i=1}^N i = 2\cdot\dfrac{N(N+1)}{2}=N(N+1)[/tex]
The second sum is simply the sum of N ones:
[tex]\underbrace{1+1+1\ldots+1}_{N\text{ times}}=N[/tex]
So, the final result is
[tex]\displaystyle \sum_{i=1}^N 2i-1 = 2\sum_{i=1}^N i - \sum_{i=1}^N 1 = N(N+1)-N = N^2+N-N = N^2[/tex]
which ends the proof.
