There are many theorems and statements in mathematics that are to be proved with the help of mathematical induction. This is a way of proving statements that those statements are true for all positive integers.

Related Calculators | |

Inductance Calculator | Inductive Reactance Calculator |

Mathematical Equation Solver | Acceleration Formula Calculator |

The technique of mathematical induction is based on the principle that if a statement is true for integers - 1, 2, 3 and so on, up to k, then it is also true for all integers.

Following steps are to be followed while

Let us consider that we are to prove that a

Step 1:

Step 3:

Step 4:

Solution:

To prove that above statement is true for n = 1

First integer = 1 whose sum will also be 1.

Therefore, it holds for n = 1

Now, we can assume that given statement is true for k positive integers

Hence, 1 + 2 + 3 + 4 + ...... + k = $\frac{k}{2}$$(k+1)$

We shall prove the validity of statement for k + 1

Sum of $k\ +\ 1$ positive integers

= 1 + 2 + 3 + 4 + .... + (k + 1)

= (1 + 2 + 3 + 4 + .... + k) + (k + 1)

We know the sum of k positive integers

= $\frac{k}{2}$$(k\ +\ 1)\ +\ (k\ +\ 1)$

= $\frac{k\ (k\ +\ 1)\ +\ 2(k\ +\ 1)}{2}$

= $\frac{(k\ +\ 1)(k\ +\ 2)}{2}$

= $(k+1)$ $\frac{[(k\ +\ 1)\ +\ 1]}{2}$

This proves that statement holds for $n$ = $k\ +\ 1$.

Therefore, we conclude that the statement holds of all n positive integers.

More topics in Mathematical Induction Formula | |

Principle of Mathematical Induction | |