Let us denote the given equality by P(n), i.e.,
\(P(n): \dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+\dfrac{1}{3.4.5}+.... \)
\( +\dfrac{1}{n(n+1)(n+2)}=\dfrac{n(n+3)}{4(n+1)(n+2)}\)
For n=1,
L.H.S. = \( \dfrac{1}{1.2.3}=\dfrac{1}{6}\)
R.H.S.= \( \dfrac{1.(1+3)}{4(1+1)(1+2)}\)
= \( \dfrac{1.4}{4.2.3}=\dfrac{1}{6}\)
Therefore, P(n) is true for n=1.
Let us assume that P(k) is true for some positive integer k , i.e.,
\( \dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+\dfrac{1}{3.4.5}+....\)
\( +\dfrac{1}{k(k+1)(k+2)}=\dfrac{k(k+3)}{4(k+1)(k+2)}\)..........(i)
Now, we have to prove that P(k+1) is also true.
Consider
\( [ \dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+\dfrac{1}{3.4.5}+....\)
\( +\dfrac{1}{k(k+1)(k+2)}]+\dfrac{1}{(k+1)(k+2)(k+3)}\)
\( \dfrac{k(k+3)}{4(k+1)(k+2)}+\dfrac{1}{(k+1)(k+2)(k+3)}\) ........[Using (i)]
\( \dfrac{1}{(k+1)(k+2)}\{\dfrac{k(k+3)}{4}+\dfrac{1}{k+3}\}\)
\( \dfrac{1}{(k+1)(k+2)}\{\dfrac{k(k+3)^2+4}{4(k+3)}\)
\( \dfrac{1}{(k+1)(k+2)}\{\dfrac{k^3+6k^2+9k+4}{4(k+3)}\}\)
\( \dfrac{1}{(k+1)(k+2)}\{\dfrac{k^3+2k^2+k+4k^2+8k+4}{4(k+3)}\}\)
\( \dfrac{1}{(k+1)(k+2)}\{\dfrac{k(k+1)^2+4(k+1)^2}{4(k+3)}\}\)
\( \dfrac{(k+1)^2(k+4)}{4(k+1)(k+2)(k+3)}\)
\( \dfrac{(k+1)\{(k+1)+3\}}{4\{(k+1)+1\}(k+1)+2\}}\)
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 Pragya Singh | 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.