Let us denote the given equality by P(n), i.e.,
P(n): \( 1.3+2.3^2+3.3^3+...+n.3^n\)
\( =\dfrac{(2n-1)3^{n+1}+3}{4}\)
For n=1,
L.H.S = 1.31=3
R.H.S \( =\dfrac{(2.1-1)3^{1+1}+3}{4}\)
\( =\dfrac{3^2+3}{4}\)
= \( \dfrac{12}{4}=3\)
Therefore, P(n) is true for n=1.
Let us assume that P(k) is true for some positive integer k , i.e.,
\( 1.3+2.3^2+3.3^3+...+k.3^k\)
\( =\dfrac{(2k-1)3^{k+1}+3}{4}\) .........(i)
Now, we have to prove that \( P(k+1)\) is also true
Consider
\( 1.3+2.3^2+3.3^3+...+k.3^k+(k+1).3^{k+1}\)
= \( 1.3+2.3^2+3.3^3+...+k.3^k+(k+1).3^{k+1}\)
= \( \dfrac{(2k-1)3^{k+1}+3}{4}+(k+1)3^{k-1}\)......... [Using (i)]
= \( \dfrac{(2k-1)3^{k+1}+3+4(k+1)3^{k+1}}{4}\)
= \( \dfrac{3^{k+1}(6k+3)+3}{4}\)
= \( \dfrac{3^{k+1}.3(2k+1)+3}{4}\)
= \( \dfrac{(3^{k+1)+1}(2k+1)+3}{4}\)
= \( \dfrac{\{2({k+1})-1\}3^{(k+1)+1}+3}{4}\)
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.