Prove by induction that the nth triangular number?

Question

Prove by induction that the nth triangular number, T (sub n), is given by T (sub n)= (n/2)(n+1).

Answer 1

The keys to induction are to show that the relationship works for the smallest integer to which it applies, then to show that if it applies for one integer, it works for the next higher one. You can then show by repeated application that the relationship works for all integers greater than or equal to the one you started with.

For this example, the first step is to show that the relationship works for n = 1. That is easy. T (sub 1) = (1/2) (1 + 1) = (1/2)(2) = 1, which is indeed the first triangular number.

For the next step, we assume that the relationship works for some value n. Because a triangular number is the sum of all integers from 1 to n, the next higher triangular number above T(sub n) plus n + 1. Substituting our expression for T(sub n), we get

T (sub n+1) = (n/2)(n+1) + n + 1
= (n/2)(n+1) + 1(n+1)
= (n/2 + 1)(n+1)
= (n/2 + 2/2)(n+1)
= (n+2) * (1/2) * (n+1)
= (n+1) * (1/2) * (n+2)
= ((n+1)/2)(n+2)
= ((n+1)/2)((n+1)+1)
which is the expected value of T (sub n+1)

By induction, since the expression for T(sub 1) holds, so does that for T(sub 2)
and because T(sub 2) holds, so does T(sub 3)
and because T(sub 3) holds, so does T(sub 4)
and so on, for any arbitrarily large integer n.

Answer 2

Base case for n=1, T(1) = 1

Now assume true for k

then

T(k+1) = T(k) + (k+1)
=(k/2)*(k+1) +(k+1)
=[(k/2)+1](k+1)
= [(k+2)/2](k+1)
={(k+1)/2](k+2)

Letting k+1 = n then you've shown that
T(n) = (n/2)(n+1)

which is what you wanted to show.

Answer 3

if n =1 T(1) = 1
if n=2 , T(2) = 2+1 = 3 = (2/2)(2+1)

Say it is correct for n

prove for n+1, T(n+1) = T(n)+(n+1) = (n/2)(n+1) + (n+1) =

(n+1)(n+2)/2 = T(n+2), QED.

Answer 4

T(1) = 1
(1/2)(1+1)=1

T(2) = 1+2 = 3
(2/2)(2+1) = 3

Let it be true for n=m
T(m)=(m/2)((m+1)
Add m+1

(m/2)(m+1) + (m+1)
=(m+1)(m/2+1)
=[(m+1)/2](m+2)

Which is the format for T(m+1)

If it is true for n=1, it must be true for n=2
If it is true for n=2, it must be true for n=3
.
.

It ha been shown to be true for n=1
Therefore it is true for all values