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)
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!
To prove which expression for Sn is true, we will utilize the method of induction.
Induction is a powerful mathematical technique that allows us to prove a statement for all positive integers (or a subset of them) by dividing it into two steps: the base case and the inductive step.
Base Case:
First, we'll prove the statement for the base case, which is usually the smallest possible input, n = 1.
For n = 1, we need to find the value of S1 and check if it satisfies any of the given expressions.
Using the formulas:
S1 = 1^2 = 1
Checking each expression:
1) S1 = 1(1-1)/3 = 0/3 = 0
2) S1 = 1(2-1)/3 = 1/3
3) S1 = 1(1+1)/3 = 2/3
4) S1 = 1(1+1)(2*1+1)/3 = 2/3
From the base case, we can see that the second, third, and fourth expressions (Sn = n(2n-1)/3, Sn = n(n+1)/3, Sn = n(n+1)(2n+1)/3) all evaluate to 1/3, meaning they satisfy the base case condition.
Inductive Step:
Next, we assume the given expression is true for some value k and try to prove it holds for the next value, k+1.
Assuming Sn = k(2k-1)/3, Sn+1 should be equal to (k+1)(2(k+1)-1)/3.
Using the induction hypothesis, we can rewrite Sn as k(2k-1)/3.
Sn+1 = Sn + (k+1)^2
= k(2k-1)/3 + (k+1)^2
Expanding and simplifying:
Sn+1 = (2k^2 - k)/3 + (k^2 + 2k + 1)
= (2k^2 - k + 3k^2 + 6k + 3)/3
= (5k^2 + 5k + 3)/3
= (k^2 + k(2k+1) + (2k+1))/3
= (k+1)(2k+1)/3
Now, we have Sn+1 in the same form as our fourth expression (Sn+1 = (k+1)(2k+1)/3).
Hence, by the principle of mathematical induction, we can conclude that the fourth expression, Sn = n(n+1)(2n+1)/3, is true for all positive integers n.