We can write the given statement as
P (n): 1.2 + 2.22 + 3.22 + … + n.2n = (n – 1) 2n+1 + 2
If n = 1 we get
P (1): 1.2 = 2 = (1 – 1) 21+1 + 2 = 0 + 2 = 2
Which is true.
Consider P (k) be true for some positive integer k
1.2 + 2.22 + 3.22 + … + k.2k = (k – 1) 2k + 1 + 2 … (i)
Now let us prove that P (k + 1) is true.
Here
1.2 + 2.22 + 3.22 + … + k.2k + (k + 1) 2k + 1
\( (k-1)2^{k+1}+2+(k+1)2^{k+1}\)
\( 2^{k+1}(k-1)+(k+1)+2\)
\( 2^{k+1}.2k+2\)
\(k.2^{(k+1)+1}+2\)
\(\{ (k+1)-1\}2^{(k+1)+1}+2\)
P (k + 1) is true whenever P (k) is true.
Therefore, by the principle of mathematical induction, statement P (n) is true for all natural numbers i.e. n.
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.