Question

Please state the steps on how to do the problem

Answer 1

Since d(1) = 1, we can assert that the induction proves that the value of n is true for n = 1, and there is only one dot. Also, the mathematical induction is true for all natural numbers (n).

What is mathematical induction?

A mathematical approach known as mathematical induction is used to demonstrate that a claim or statement is true for each and every natural number.

You must demonstrate the following proposition using an inductive step of proof such that d(k) and d(k+1) are both true if d(k) is true.

To Prove:

$\mathbf{d(n) = \dfrac{n(n+1)}{2}}$

For n = 1

$\mathbf{d(1) = \dfrac{1(1+1)}{2}}$

$\mathbf{d(1) = \dfrac{1 \times 2}{2}}$

$\mathbf{d(1) = \dfrac{2}{2}}$

d(1) = 1   this is true for n = 1, there is only one dot.

b.

Now, let us assume that the result is true for n = k + 1 such that:

$\mathbf{d(k+1) = \dfrac{(k+1) ((k+1) + 1) }{2}}$

$\mathbf{d(k+1) = \dfrac{(k+1) (k+2) }{2} ---- (1)}$

Now, Let n = k, then:

$\mathbf{d(k) = \dfrac{k(k + 1)}{2} ----(2)}$

Recall that, the total number of dots d(n) increases by 'n' each time.

i.e. d(n+1) = d(n) + n

Therefore;

d(k) = d(k+1) + k

To prove:

• d(k) + (k+1) = d(k+1), we have:

$\mathbf{\dfrac{k(k + 1)}{2} + (k + 1) = \dfrac{(k+1) (k+2) }{2}}$

$\mathbf{ \dfrac{k(k+1)+ (2k+2) }{2}= \dfrac{(k+1) (k+2) }{2}}$

$\mathbf{ \dfrac{(k+1) (k+2) }{2}= \dfrac{(k+1) (k+2) }{2}}$

L.H.S    =   R.H.S

Therefore, we can conclude that the mathematical induction is true for all natural numbers (n).

Learn more about how to use induction to prove here:

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