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.