\( 1.2+2.3+3.4+...+n(n+1) =[\dfrac{n(n+1)(n+2)}{3}]\)
For n=1,
L.H.S.=1.2=2
R.H.S. \( =\dfrac{1(1+1)(1+2)}{3}\)
= \( \dfrac{1.2.3}{3}=2\)
Therefore, P(n) is true for n=1.
Let us assume that P(k) is true for some positive integer k , i.e.,
\( 1.2+2.3+3.4+...+k(k+1) \)
\( =[\dfrac{k(k+1)(k+2)}{3}]\) ...........(i)
Now, we have to prove that P (k+1) is also true.
Consider
\( 1.2+2.3+3.4+...+k(k+1) \)
\( +(k+1)(k+2)\)
\( =\dfrac{k(k+1)(k+2)}{3}+(k+1)(k+2)\) [using (i)]
\(= (k+1)(k+2)(\dfrac{k}{3}+1)\)
= \( \dfrac{(k+1)(k+2)(k+3)}{3}\)
\(= \dfrac{(k+1)(k+1+1)(k+1+2)}{3}\)
Therefore, P(k+1) holds whenever P(k)holds.
Hence, the given equality is true for all natural numbers i.e., N by the principle of mathematical induction
Answered by Abhisek | 1 year agoGiven an example of a statement P (n) such that it is true for all n ϵ N.
a + (a + d) + (a + 2d) + … + (a + (n-1)d) = \( \dfrac{n}{2}\) [2a + (n-1)d]
72n + 23n – 3. 3n – 1 is divisible by 25 for all n ϵ N
n (n + 1) (n + 5) is a multiple of 3 for all n ϵ N.