use mathematical induction to prove that 1^2 + 2^2 + 3^2 + ... + n^2 = n(n+1)(2n+1)/6 for all positive integral values of n.
My work: n=1: 1(1+1)(2(1)+1)/6 = 1(2)(3)/6 = 6/6 = 1.
1^2+2^2+3^2...(n-1)^2+n^2= [(n-1)(n-1)+2n(n-1)+1]/6 = 2n^3+3n^2+n/6 = n(2n=1)(n+1)/6 = n(n+1)(2n+1)/6. Since Sn is valid for n=1, it is also valid for n=2, n=3, etc.
The teacher marked this as wrong but this followed the example provided by the book. Where did I go wrong?
Step 1
show it to be true for n=1
LS = 1^2 = 1
RS = (1)(2)(3)/6 = 1, checks out
Step 2
Assume it is true for n=k
that is..
1^2 + 2^2 + 3^2 + ... + k^2 = k(k+1)(2k+1)/6
Step 3
is it then true for n = k+1 ??
or
is 1^2 + 2^2 + 3^2 + ... + k^2 + (k+1)^2 = (k+1)(k+2)(2k+3)/6 ??
from Step 2 the sum of terms to the end of k^2 = k(k+1)(2k+1)/6
then
LS = k(k+1)(2k+1)/6 + (k+1)^2
= (k(k+1)(2k+1) + 6(k+11)^2)/6
= (k+1)(k(2k+1) + 6(k+1))/6
= (k+1)(2k^2 + k + 6k + 6)/6
= (k+1)(k+2)(2k+3)/6
= RS
Q.E.D.
compare your solution to mine.
Where you not shown the 3step process of an induction proof?
To use mathematical induction, you need to follow a standardized process. Let's go through the steps to see where you might have gone wrong:
Step 1: Base Case
You correctly started by showing that the equation holds true for n = 1. This is the base case and serves as the starting point for the induction proof.
Step 2: Inductive Hypothesis
Assume that the equation holds true for some arbitrary positive integer k, where k is greater than or equal to 1. In other words, assume that
1^2 + 2^2 + 3^2 + ... + k^2 = k(k+1)(2k+1)/6.
This assumption is called the "inductive hypothesis."
Step 3: Inductive Step
Now, you need to prove that if the equation holds true for k, then it also holds true for k+1. This is the critical step that you missed in your explanation.
To do this, start by adding (k+1)^2 to both sides of the equation assumed in the inductive hypothesis:
1^2 + 2^2 + 3^2 + ... + k^2 + (k+1)^2 = k(k+1)(2k+1)/6 + (k+1)^2.
Simplifying the right side, we have:
[(k+1)(k(2k+1)/6 + (k+1)] = [(k+1)(2k^2+k)/6 + (k+1)^2] = (k+1)(2k^2+k + 6k^2 +6k)/6.
Combining the terms and simplifying further, we get:
(k+1)(2k^2+k+3k^2+3k)/6 = (k+1)(5k^2 + 4k)/6 = (k+1)(k+2)(2k+1)/6.
This expression is in the form of the equation we want to prove, but with k replaced by k+1. So, the equation holds true for k+1 as well.
Step 4: Conclusion
Since the equation holds true for the base case (n = 1) and has been shown to hold true for k implies k+1, we can conclude that the equation holds true for all positive integral values of n by mathematical induction.
In your explanation, you correctly evaluated the equation for n = 1 but failed to demonstrate the inductive step where you show that the equation holds for (n-1) and add n^2 to both sides, proving it for n as well (which involves writing out the equation for (n-1)^2 and then simplifying). Without this inductive step, your proof is incomplete.