Let us denote the given equality by P(n), i.e.,
\(P(n): \dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}...+\dfrac{1}{2^n}=1-\dfrac{1}{2^n}\)
For n=1,
L.H.S.= \( \dfrac{1}{2}\)
R.H.S.\(1- \dfrac{1}{2^1}=\dfrac{1}{2}\)
Therefore, P(n) is true for n=1.
Let us assume that P(k) is true for some positive integer k , i.e.,
\(\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}...+\dfrac{1}{2^k}=1-\dfrac{1}{2^k}\).........(i)
Now, we have to prove that P(k+1) is also true.
Consider
\(( \dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}...+\dfrac{1}{2^k})+\dfrac{1}{2^{k+1}}\)
\( 1-\dfrac{1}{2^k}+\dfrac{1}{2^{k+1}}\)............[Using (i)]
\( 1-\dfrac{1}{2^k}( 1-\dfrac{1}{2})\)
\( 1-\dfrac{1}{2^k}( \dfrac{1}{2})\)
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 Pragya Singh | 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.