Question

Prove by mathematical induction that 1^2 + 2^2 +3^2...+k^2= k(k+1)(2k+1)/6​

Answer 1

As usual, we start by checking that the base case holds: if
`k=1` we have

`1^2 = (1\cdot 2 \cdot 3)/(6)`

which is true. Now, we assume that

`1^2+2^2+\ldots+k^2=(k(k+1)(2k+1))/(6)`

and we check
`P(k+1)`: if we add the k+1-th term on both sides, we have

`1^2+2^2+\ldots+k^2+(k+1)^2=(k(k+1)(2k+1))/(6)+(k+1)^2`

And our goal is to rearrange the right hand term as follows:

`(k(k+1)(2k+1))/(6)+(k+1)^2 = (2k^3+3k^2+k)/(6)+(6k^2+12k+6)/(6)`

`=(2k^3+9k^2+13k+6)/(6) = ((2k+3)(k+1)(k+2))/(6)`

And we're done, because
`(2k+3)(k+1)(k+2)` is exactly the formula
`k(k+1)(2k+1)` if you substitute k with k+1