# Mathematical Induction

## Mathematical Induction

Principle of finite Mathematical Induction: Let {S (n)/ n ϵ N} be a set of statements.

Step 1: Let S (n) is given statement, S (1) is true.

Step 2: Assume that the statement S (m) is true.

Step 3: Assume S (m) is true ⇔ S (m + 1) statement is true.

Example: Prove by mathematical induction $${{1}^{3}}+{{2}^{3}}+{{3}^{3}}+……..+{{n}^{3}}=\frac{{{n}^{2}}{{(n+1)}^{2}}}{4}$$.

Solution: Let S(n) is be the statement.

$$S(n)={{1}^{3}}+{{2}^{3}}+{{3}^{3}}+……..+{{n}^{3}}=\frac{{{n}^{2}}{{(n+1)}^{2}}}{4}$$,

Step 1: Let S (n) is given statement, S (1) is true for n = 1

L.H.S n³ = 1

And R.H.S $$\frac{{{n}^{2}}{{(n+1)}^{2}}}{4}$$.

Put n = 1

$$\frac{{{1}^{2}}{{(1+1)}^{2}}}{4}=1$$.

L.H.S = R.H.S

Therefore S (1) is true

Step 2: Let assume S(m) is true

$$S(m)={{1}^{3}}+{{2}^{3}}+{{3}^{3}}+……+{{m}^{3}}=\frac{{{m}^{2}}{{(m+1)}^{2}}}{4}$$ … (1)

Step 3: Assume S(m) is true ⇔ S (m + 1) statement is true

$$S(m+1)={{1}^{3}}+{{2}^{3}}+{{3}^{3}}+……+{{(1+m)}^{3}}=\frac{{{(1+m)}^{2}}{{(m+1+1)}^{2}}}{4}$$ … (2)

From equation (1) Adding terms on both sides

$$S(m)={{1}^{3}}+{{2}^{3}}+{{3}^{3}}+……+{{m}^{3}}+{{(1+m)}^{3}}=\frac{{{m}^{2}}{{(m+1)}^{2}}}{4}+{{(1+m)}^{3}}$$,

= $$\frac{{{m}^{2}}{{(m+1)}^{2}}}{4}+{{(1+m)}^{3}}$$,

= $$\frac{{{m}^{2}}{{(1+m)}^{2}}+4{{(1+m)}^{3}}}{4}$$,

= $$\frac{{{(1+m)}^{2}}({{m}^{2}}+4+4m)}{4}$$,

= $$\frac{{{(1+m)}^{2}}{{(m+2)}^{2}}}{4}$$,

= $$\frac{{{(1+m)}^{2}}{{(m+1+1)}^{2}}}{4}$$,

S (m +1) is true

Therefore S (m) is true.