Answer
This proof follows the basic pattern of the solution to Exercise 3, but the algebra gets more complex. The
statement P( n) that we wish to prove is
1
4 2
4 3
4 4 n(n+1)(2n+l)(3n2 +3n-l) + + + · .. +n = 30 '
where n is a positive integer. The basis step, n = 1, is true, since 1 · 2 · 3 · 5/30 = 1. Assume the displayed
statement as the inductive hypothesis, and proceed as follows to prove P(n + 1):
(14 24 4) ( l) 4 n(n+l)(2n+1)(3n2 +3n-l) ( )4 + +· .. +n + n+ = + n+l 30
n+l = 3() (n(2n + 1)(3n2 + 3n - 1) + 30(n + 1) 3
)
n+l 4 3 2 = ----W-(6n + 39n + 9ln + 89n + 30)
n+l = ----W-(n + 2)(2n + 3)(3(n + 1) 2 + 3(n + 1) -1)
The last equality is straightforward to check; it was obtained not by attempting to factor the next to last
expression from scratch but rather by knowing exactly what we expected the simplified expression to be.
Work Step by Step
This proof follows the basic pattern of the solution to Exercise 3, but the algebra gets more complex. The
statement P( n) that we wish to prove is
1
4 2
4 3
4 4 n(n+1)(2n+l)(3n2 +3n-l) + + + · .. +n = 30 '
where n is a positive integer. The basis step, n = 1, is true, since 1 · 2 · 3 · 5/30 = 1. Assume the displayed
statement as the inductive hypothesis, and proceed as follows to prove P(n + 1):
(14 24 4) ( l) 4 n(n+l)(2n+1)(3n2 +3n-l) ( )4 + +· .. +n + n+ = + n+l 30
n+l = 3() (n(2n + 1)(3n2 + 3n - 1) + 30(n + 1) 3
)
n+l 4 3 2 = ----W-(6n + 39n + 9ln + 89n + 30)
n+l = ----W-(n + 2)(2n + 3)(3(n + 1) 2 + 3(n + 1) -1)
The last equality is straightforward to check; it was obtained not by attempting to factor the next to last
expression from scratch but rather by knowing exactly what we expected the simplified expression to be.
