Consider :\(P(k) : \sum _{ t=1 }^{ k-1 }{ t(t+1) } =\frac { k(k-1)(k+1) }{ 3 } ​​​​k\ge 2\) 
P(k):1.2+2.3+3.4+....+(k-1)k \(=\frac { k(k-1)(k+1) }{ 3 } \)
Now P(k+1):1.2+2.3+3.4+...+(k1)k+k(k+1)
\(=\frac { k(k-1)(k+1) }{ 3 } +k(k+1)\)
\(=\frac { k(k+1)(k+2) }{ 3 } k\ge 2\)

Consider  P(k):2^2k -1= \(3\lambda \) (say)
Now P(k+1):2^2k+1 -1 = 2^2k.2² -1
\((3\lambda +1)4-1=12\lambda +3\)
\(3(4\lambda +1)\), which is divisible by 3.

\(\frac { 1 }{ 2.5 } +\frac { 1 }{ 5.8 } +\frac { 1 }{ 8.11 } +....+\frac { 1 }{ (3k-1)(3k+2) } =\frac { k }{ 6k+4 } \)
Now P(k+1):\(\frac { 1 }{ 2.5 } +\frac { 1 }{ 5.8 } +\frac { 1 }{ 8.11 } +....+\frac { 1 }{ (3k-1)(3k+2) } +\frac { 1 }{ [3(k+1)-1][3(k+1)+2] } \)
=\(\frac { k }{ 2(3k+2) } +\frac { 1 }{ (3k+2)(3k+5) } \)
=\(\frac { { 3k }^{ 2 }+5k+2 }{ 2(3k+2)(3k+5) } +\frac { (3k+2)(k+1) }{ 2(3k+2)(3k+5) } =\frac { k+1 }{ 2(3k+5) } \)
=\(\frac { k+1 }{ 6(k+1)+4 } \)

Consider 
P(k):\(\left( 1+\frac { 1 }{ 1 } \right) \left( 1+\frac { 1 }{ 2 } \right) \left( 1+\frac { 1 }{ 3 } \right) .....\left( 1+\frac { 1 }{ k } \right) =k+1\)
P(k+1):\(\left[ \left( 1+\frac { 1 }{ 1 } \right) \left( 1+\frac { 1 }{ 2 } \right) \left( 1+\frac { 1 }{ 3 } \right) .....\left( 1+\frac { 1 }{ k } \right) \right] \left( 1+\frac { 1 }{ k+1 } \right) \)
\(=(k+1)\left(\frac{k+2}{k+1}\right) \)
=(k+1)+1

Consider P(k) : 1+2+2²+.... +2^k = 2^k+1-1
Now P(+1):1+2+2² +..+2^k = 2^k+1 
=2^k+1-1+2^k+1
=2^(k+1)+1 -1

Consider P(k):2+2.3+2.3² + ....+2.3^k-1 = (3^k-1)
Now,P(k+1):2+2.3+2.3² + ....+2.3^k-1 +2.3^k
=(3^K -1)+2.3^k =3.3^k -1 =3^k+1-1

Consider P(k):2+4+6+8+.....+2k = k(k+1)
Now, P(k+1):2+4+6+.....2(k)+2(k+1)
=k(k+1)+2(k+1)=k² + 3k+2
=(k+1)(k+1)

Consider P(k):2+4+6+8+.....+2k = k(k+1)
Now, P(k+1):2+4+6+.....2(k)+2(k+1)
=k(k+1)+2(k+1)=k² + 3k+2
=(k+1)(k+1)

Let P(n): 2ⁿ > n
When n =1, 2¹ >1. Hence P(1) is true.
Assume that P(k) is true for any positive integer k, i.e.,
2^k > k ... (1)
We shall now prove that P(k +1) is true whenever P(k) is true.
Multiplying both sides of (1) by 2, we get
2. 2^k > 2k
i.e., \(2^{k+1}>2 k=k+k>k+1\)
Therefore, P(k + 1) is true when P(k) is true. Hence, by principle of mathematical induction, P(n) is true for every positive integer n.
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