-
Problem 1: Prove that for every positive integer n,
\[
\sum_{j=1}^{n} j(j+1)(j+2) = \frac{1}{4}n(n+1)(n+2)(n+3)
\]
Proof by Mathematical Induction:
Base Case (n = 1):
Left-hand side: \(1 \cdot 2 \cdot 3 = 6\)
Right-hand side: \(\frac{1}{4} \cdot 1 \cdot 2 \cdot 3 \cdot 4 = \frac{24}{4} = 6\)
Both sides are equal. Base case holds.
Inductive Step:
Assume the formula holds for some positive integer \(k\), i.e.,
\[
\sum_{j=1}^{k} j(j+1)(j+2) = \frac{1}{4}k(k+1)(k+2)(k+3)
\]
We need to show it holds for \(k+1\):
\[
\sum_{j=1}^{k+1} j(j+1)(j+2) = \frac{1}{4}(k+1)(k+2)(k+3)(k+4)
\]
Starting from the left-hand side:
\[
\sum_{j=1}^{k+1} j(j+1)(j+2) = \sum_{j=1}^{k} j(j+1)(j+2) + (k+1)(k+2)(k+3)
\]
By the inductive hypothesis:
\[
= \frac{1}{4}k(k+1)(k+2)(k+3) + (k+1)(k+2)(k+3)
\]
Factor out \((k+1)(k+2)(k+3)\):
\[
= (k+1)(k+2)(k+3) \left( \frac{k}{4} + 1 \right) = (k+1)(k+2)(k+3) \left( \frac{k+4}{4} \right)
\]
\[
= \frac{1}{4}(k+1)(k+2)(k+3)(k+4)
\]
This matches the right-hand side for \(n = k+1\).
Therefore, by mathematical induction, the formula holds for all positive integers \(n\).
-
Problem 2: Prove that 6 divides \(n^3 - n\) whenever \(n\) is a non-negative integer.
Proof by Mathematical Induction:
Base Case (n = 0):
\(0^3 - 0 = 0\), and 6 divides 0. Base case holds.
Inductive Step:
Assume 6 divides \(k^3 - k\) for some non-negative integer \(k\), i.e., \(k^3 - k = 6m\) for some integer \(m\).
We need to show 6 divides \((k+1)^3 - (k+1)\).
Expand:
\[
(k+1)^3 - (k+1) = k^3 + 3k^2 + 3k + 1 - k - 1 = k^3 + 3k^2 + 2k
\]
Rearrange:
\[
= (k^3 - k) + 3k^2 + 3k = (k^3 - k) + 3k(k + 1)
\]
By the inductive hypothesis, \(k^3 - k\) is divisible by 6.
Now consider \(3k(k+1)\). Since \(k\) and \(k+1\) are consecutive integers, one of them is even, so \(k(k+1)\) is even. Thus, \(3k(k+1)\) is divisible by \(3 \cdot 2 = 6\).
Therefore, \((k+1)^3 - (k+1)\) is the sum of two terms each divisible by 6, so it is also divisible by 6.
Hence, by mathematical induction, 6 divides \(n^3 - n\) for all non-negative integers \(n\).
*(Alternative direct proof: \(n^3 - n = n(n-1)(n+1)\), which is the product of three consecutive integers. Among any three consecutive integers, at least one is divisible by 2 and exactly one is divisible by 3, so the product is divisible by 6.)*
Parent Tip: Review the logic above to help your child master the concept of mathematical induction worksheet.
