Question

what is the value of summation of (1/2)to the power n where n=0 to infinity​

Answer 1

Answer:

2

Step-by-step explanation:

Given :

To find the value of :

  ∞

  Σ $(\frac{1}{2})^n$ = $1 + \frac{1}{2} + \frac{1}{4} + ...$∞

n = 0

Sum of ∞ terms in G.P is given by

⇒ $S_n = \frac{a}{1-r}$

Where a = first term = 1, r = common ratio = $\frac{1}{2}$

⇒ $S_n = \frac{1}{1 - \frac{1}{2}} = \frac{1}{(\frac{1}{2})} = 2$

Logically,

If n = 1,

$1 + \frac{1}{2} = \frac{3}{2} = \frac{2^2 - 1}{2^1}$,

If n = 2,

$1 + \frac{1}{2} + \frac{1}{4} = \frac{3}{2} + \frac{1}{4} = \frac{7}{4} = \frac{2^3 - 1}{2^2}$,

If n = 3,

$1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} = \frac{3}{2} + \frac{1}{4} + \frac{1}{8} = \frac{7}{4} \frac{1}{8} = \frac{15}{8} = = \frac{2^4 - 1}{2^3}$,

The values are getting very close to 2,

⇒ $1 + \frac{1}{2} + \frac{1}{4} + ... +(\frac{1}{2})^{n} =\frac{2^{n+1} - 1}{2^n}$

⇒$1 + \frac{1}{2} + \frac{1}{4} + ...(\frac{1}{2})^{\infty} =\lim_{x \to 2} x = 2$