Question

What is the sum of the first 70 consecutive odd numbers? Explain.

Answer 1

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

$\displaystyle \sum_{i=1}^N 2i-1 = 2\sum_{i=1}^N i - \sum_{i=1}^N 1$

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

$\displaystyle 2\sum_{i=1}^N i = 2\cdot\dfrac{N(N+1)}{2}=N(N+1)$

The second sum is simply the sum of N ones:

$\underbrace{1+1+1\ldots+1}_{N\text{ times}}=N$

So, the final result is

$\displaystyle \sum_{i=1}^N 2i-1 = 2\sum_{i=1}^N i - \sum_{i=1}^N 1 = N(N+1)-N = N^2+N-N = N^2$

which ends the proof.