Let us denote the given equality by P(n), i.e.,
\( \dfrac{1}{1.4}+ \dfrac{1}{4.7}+ \dfrac{1}{7.10}+...\)
\( +\dfrac{1}{(3n-2)(3n+1)}=\dfrac{n}{(3n+1)}\)
For n=1,
L.H.S.= \( \dfrac{1}{1.4}\)= \( \dfrac{1}{4}\)
R.H.S.= \( \dfrac{1}{3.1+1}= \dfrac{1}{4}\)
Therefore, P(n) is true for n=1.
Let us assume that P(k) is true for some positive integer k , i.e.,
\( \dfrac{1}{1.4}+ \dfrac{1}{4.7}+ \dfrac{1}{7.10}+...\)
\( +\dfrac{1}{(3k-2)(3k+1)}=\dfrac{k}{(3k+1)}\).......(i)
Now, we have to prove that P(k+1) is also true.
Consider
\(\{ \dfrac{1}{1.4}+ \dfrac{1}{4.7}+ \dfrac{1}{7.10}+...\)
\( +\dfrac{1}{(3k-2)(3k+1)}\}\)
\( +\dfrac{1}{\{3(k+1)-2\}\{3(k+1)+1\}}\)
\(\dfrac{k}{3k+1} +\dfrac{1}{(3k+1)(3k+14)}\) [Using (i)]
\( \dfrac{1}{3k+1} \{k +\dfrac{1}{(3k+4)}\}\)
\( \dfrac{1}{(3k+1)} \{\dfrac{3k^2+4k+1}{(3k+4)}\}\)
\( \dfrac{1}{(3k+1)} \{\dfrac{3k^2+3k+k+1}{(3k+4)}\}\)
\( \dfrac{(3k+1)(k+1)}{(3k+1)(3k+4)}\)
\( \dfrac{(k+1)}{3(k+1)+1} \)
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.