Let us denote the given equality by P(n), i.e.,
\( 1.3+3.5+5.7+....+(2n-1)(2n+1) \)
\( =\dfrac{n(4n^2+6n-1)}{3}\)
For n=1,
L.H.S. = 1.3 = 3
\(R.H.S =\dfrac{1(4.1^2+6.1-1)}{3} \)
\( =\dfrac{n(4+6-1)}{3}\)
\( =\dfrac{9}{3}=3\)
Therefore, P(n) is true for n=1.
Let us assume that P(k) is true for some positive integer k , i.e.,
\( 1.3+3.5+5.7+....+(2k-1)(2k+1) \)
\( =\dfrac{k(4k^2+6k-1)}{3}\).........(i)
Now, we have to prove that P( k+1) is also true.
Consider
\( 1.3+3.5+5.7+....+(2k-1)(2k+1) \)
\( +\{(k+1)-1\}\{2(k+1)+1\}\)
\(\dfrac{k(4k^2+6k-1)}{3}+(2k+2-1)(2k+2+1)\)
\( \dfrac{k(4k^2+6k-1)}{3}+(2k+1)(2k+3)\)
\( \dfrac{k(4k^2+6k-1)}{3}+(4k^2+8k+3)\)
\( \dfrac{k(4k^2+6k-1)+3(4k^2+8k+3)}{3}\)
= \(\dfrac{4k^2+18k^2+23k+9}{3}\)
= \( \dfrac{4k^2+14k^2+9k+4k^2+14k+9}{3}\)
= \( \dfrac{k(4k^2+14k^2+9)+1(4k^2+14k+9)}{3}\)
= \( \dfrac{(k+1)\{4(k^2+2k+1)+6(k+1)-1\}}{3}\)
= \( \dfrac{(k+1)\{4(k+1)^2+6(k+1)-1\}}{3}\)
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.