since you are adding up n squares, the sum will be something on the order of n*n^2, or n^3. So, the first 3 choices can be eliminated right away.
Going with the last one, we see right off there's a typo.
S1 = 1 = 1(2)(3)/6
Now, assume that
1+4+9...+k^2 = k(k+1)(2k+1)/6
adding in (k+1)^2 to both sides, we get
1+4+9...+k^2+(k+1)^2 = k(k+1)(2k+1)/6 + (k+1)^2
= (2k^3+3k^2+k)/6 + (k^2+2k+1)
= (2k^3 + 3k^2 + k + 6k^2 + 12k + 6)/6
= (2k^3+6k^2+6k+2 + 3k^2+6k+3 + k+1)/6
= (2(k+1)^3 + 3(k+1)^2 + (k+1))/6
ta-da!
If Sn represents the sum of the squares of the first n natural numbers, use proof by induction to find which of the following expressions for Sn is true?
Sn=n(n-1)/(3)
Sn=n(2n-1)/(3)
Sn=n(n+1)/(3)
Sn=n(n+1)(2n+1)/(3)
1 answer
1 answer