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Sir please provide solution for this question 
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Asked by Surjitsp6u | 25 Jul, 2019, 11:23: AM
answered-by-expert Expert Answer
sum from k equals 0 to n of space left parenthesis k plus 1 right parenthesis space integral subscript 0 superscript 1 space 2 to the power of negative left parenthesis k plus 1 right parenthesis x end exponent space d x equals sum from k equals 0 to n of space integral subscript 0 superscript 1 space left parenthesis k plus 1 right parenthesis space 2 to the power of negative left parenthesis k plus 1 right parenthesis x end exponent space d x T a k e space left parenthesis k plus 1 right parenthesis x equals t space rightwards double arrow space left parenthesis k plus 1 right parenthesis d x equals d t W h e n space x equals 0 comma space t equals 0 space a n d space x equals 1 comma space t equals k plus 1 equals sum from k equals 0 to n of space integral subscript 0 superscript k plus 1 end superscript space space 2 to the power of negative t end exponent space d t equals sum from k equals 0 to n of space space open square brackets negative fraction numerator begin display style 2 to the power of t end style over denominator begin display style ln 2 end style end fraction close square brackets subscript 0 superscript negative left parenthesis k plus 1 right parenthesis end superscript equals negative fraction numerator 1 over denominator ln begin display style 2 end style end fraction open square brackets sum from k equals 0 to n of open parentheses 1 over 2 to the power of k plus 1 end exponent close parentheses minus n close square brackets equals negative fraction numerator begin display style 1 end style over denominator begin display style ln 2 end style end fraction open square brackets 1 half plus 1 over 2 squared plus.. plus 1 over 2 to the power of n minus n close square brackets equals negative fraction numerator begin display style 1 end style over denominator begin display style ln 2 end style end fraction open square brackets fraction numerator begin display style 1 half open parentheses 1 minus 1 over 2 to the power of n close parentheses end style over denominator 1 minus begin display style 1 half end style end fraction minus n close square brackets equals negative fraction numerator begin display style 1 end style over denominator begin display style ln 2 end style end fraction open square brackets 1 minus 1 over 2 to the power of n minus n close square brackets T h e r e f o r e comma space sum from n equals 0 to infinity of space fraction numerator 1 over denominator n factorial end fraction space open square brackets sum from k equals 0 to n of space left parenthesis k plus 1 right parenthesis space integral subscript 0 superscript 1 space 2 to the power of negative left parenthesis k plus 1 right parenthesis x end exponent space d x close square brackets space equals negative fraction numerator begin display style 1 end style over denominator begin display style ln 2 end style end fraction space sum from k equals 0 to infinity of space fraction numerator 1 over denominator n factorial end fraction space open parentheses open square brackets 1 minus 1 over 2 to the power of n minus n close square brackets close parentheses N o w comma space f i n d space t h e space s u m.
Answered by Renu Varma | 01 Aug, 2019, 15:32: PM

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