calculate-the-value-of-n-0-1947-1-2-n-2-1947-

Question Number 62669 by Sayantan chakraborty last updated on 24/Jun/19

c a l c u l a t e t h e v a l u e o f \underset{n = 0}{\sum^{1947}} \frac{1}{2^{n} + \sqrt{2^{1947}}}

Answered by tanmay last updated on 24/Jun/19

a = \sqrt{2^{1947}}

S = \frac{1}{2^{0} + a} + \frac{1}{2^{1} + a} + \frac{1}{2^{2} + a} + \frac{1}{2^{3} + a} + \dots + \frac{1}{2^{1947} + a}

\frac{2 + a}{2} ⩾ {(2 \times a)}^{\frac{1}{2}}

2 + a ⩾ 2 \times {(2 a)}^{\frac{1}{2}}

\frac{1}{2 + a} ⩽ \frac{1}{2 \times {(2 a)}^{\frac{1}{2}}}

\frac{1}{2 + a} ⩽ \frac{1}{2 \sqrt{a} \times {(2)}^{\frac{1}{2}}}

S ⩽ \frac{1}{2 \sqrt{a}} [\frac{1}{{(2^{0})}^{\frac{1}{2}}} + \frac{1}{{(2^{1})}^{\frac{1}{2}}} + \frac{1}{{(2^{2})}^{\frac{1}{2}}} + \dots + \frac{1}{{(2^{1947})}^{\frac{1}{2}}}]

S ⩽ \frac{1}{2 \sqrt{a}} [\frac{1}{{(\sqrt{2})}^{0}} + \frac{1}{{(\sqrt{2})}^{1}} + \frac{1}{{(\sqrt{2})}^{2}} + \dots + \frac{1}{{(\sqrt{2})}^{1947}}]

S ⩽ \frac{1}{2 \sqrt{a}} \times \frac{\frac{1}{{(\sqrt{2})}^{0}} {1 - {(\frac{1}{\sqrt{2}})}^{1948}}}{1 - (\frac{1}{\sqrt{2}})} [u s i n g \frac{A (1 - r^{n})}{1 - r}]

S ⩽ \frac{1}{2 \sqrt{a}} \times \frac{\sqrt{2} \times (1 - \frac{1}{2^{974}})}{\sqrt{2} - 1}

S ⩽ \frac{1}{(2 - \sqrt{2})} \times \frac{1 - \frac{1}{2^{974}}}{{(\sqrt{2^{1947}})}^{\frac{1}{2}}} \approx \frac{1}{(2 - \sqrt{2})} \times \frac{1}{{(\sqrt{2^{1947}})}^{\frac{1}{2}}}

i h a v e f o u n d a p p r o x i m a t e v a l u e n o t

e x a c t v a l u e s o p l e a s e c h e a c k

Commented by Sayantan chakraborty last updated on 25/Jun/19

a n s w e r i s \frac{487}{\sqrt{} 2^{1945}} .

Answered by ajfour last updated on 30/Jun/19

S = \underset{n = 0}{\sum^{2 N}} \frac{1}{2^{n} + 2^{N}} (w h e r e 2 N = 1947)

= \frac{1}{2^{N}} (\underset{n = 0}{\sum^{2 N}} \frac{1}{1 + 2^{n - N}})

2^{N} S = \frac{2^{N}}{2^{N} + 1} + \frac{2^{N}}{2^{N} + 2} + \frac{2^{N}}{2^{N} + 2^{2}} + \dots

\dots \dots + \frac{2}{2 + 1} + \frac{1}{1 + 1} + \frac{1}{1 + 2} + \frac{1}{1 + 2^{2}} + . .

\dots . . + \frac{2^{2}}{2^{2} + 2^{N} +} \frac{2}{2 + 2^{N}} + \frac{1}{1 + 2^{N}} + \dots .

\Rightarrow 2^{N} S = \frac{1}{2} + N

S = \frac{2 N + 1}{2 \sqrt{2^{2 N}}}

\Rightarrow S = \frac{1948}{2 (\sqrt{} 2^{1947})} = \frac{487}{\sqrt{} 2^{1945}} .

Commented by Sayantan chakraborty last updated on 30/Jun/19

\boldsymbol Y e s \boldsymbol i t \boldsymbol i s \boldsymbol c o r r e c t \boldsymbol s o l u t i o n .

Y e s i t i s c o r r e c t s o l u t i o n .