Please send the complete solution showing all the steps.

Asked by sudhanshubhushanroy | 20th Nov, 2017, 08:20: AM

Expert Answer:

begin mathsize 16px style generalizing comma considering space straight n space terms
sum from straight r equals 1 to straight n of fraction numerator straight C presuperscript straight n subscript left parenthesis 2 straight r minus 1 right parenthesis end subscript over denominator 2 straight r end fraction
equals fraction numerator sum from straight r equals 1 to straight n of open parentheses straight C presuperscript straight n subscript left parenthesis 2 straight r minus 1 right parenthesis end subscript close parentheses over denominator sum from straight r equals 1 to straight n of open parentheses 2 straight r close parentheses end fraction
equals fraction numerator 2 to the power of straight n minus 1 end exponent over denominator straight n left parenthesis straight n plus 1 right parenthesis end fraction...... sum space of space odd space terms space in space binomials end style

Answered by Arun | 29th Nov, 2017, 11:41: AM