What is the answer? 4b/3b3
2 answers:
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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
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
The second sum is simply the sum of N ones:
So, the final result is
which ends the proof.