Prove the following by using the principle of mathematical induction for all \( n\in N\)

\( 1.2+2.2^2+3.2^2+...+n.2^n=(n-1)2^{n+1}+2\)

Asked by Pragya Singh | 1 year ago |  58

1 Answer

Solution :-

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 ago

Related Questions

a + (a + d) + (a + 2d) + … + (a + (n-1)d) = \( \dfrac{n}{2}\) [2a + (n-1)d]

Class 11 Maths Principle of Mathematical Induction View Answer