Question

The sum of the first ……………………. odd natural number is n2​

Answer 1

Answer:

Prove that the sum of the n first odd positive integers is n2, i.e., 1 + 3 + 5 + ··· + (2n − 1) = n2. Answer: Let S(n)=1+3+5+ ··· + (2n − 1).

Step-by-step explanation:

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Answer 2

Answer:

You could draw a grid and it'll all seem obvious, but then I wouldn't have to chance to show off... MATHEMATICAL INDUCTION!! Which is a lot longer and less elegant, but I think more definitive.

Let's examine our very first case. The number 1. We can easily verify that this theory holds.

1=12 .

Assume that the theory holds for the nth odd natural number (2n−1) . That is to say,

1+3+5+...+(2n−1)=(2n−1)2

We're just assuming here! We don't actually know if it's true! All we know is that it works when n=1

Our next natural odd number is (2(n+1)−1)=2n+1 What is the sum of our first (n+1) odd numbers, that is to say:

1+3+5+...+(2n−1)+(2n+1)

Well, let's just go ahead and assume that what we're trying to prove is true. Which is to say that

we can replace 1+3+5+...+(2n−1)

with n2

Using the above replacement/substitution, we get:

1+3+5+...+(2n−1)+(2n+1)=n2+(2n+1) .

Assuming what we're trying to prove is true, we get:

1+3+5+...+(2n+1)=(n+1)2

This is exactly what our original theory is, except we replace n with (n+1).

Are the two equal? A little algebraic expansion yields that:

(n+1)2=n2+2n+1 .

Thus:

- If we assume that the theory is true for any arbitrary odd natural number, it must be true for the next odd natural number.

- We know that the theory is true for the first odd natural number. If you apply the above result forever (i.e. "by mathematical induction", you can conclude that the theory is true for all odd natural numbers.