The coeffiucient of x^k ( 0<=k<= n )in the expansion of E = 1 + (1+x) + (1 + x)^2 + ........ + ( 1 + x )^n is ... . Calculate the same.

Asked by sudhanshubhushanroy | 6th Nov, 2017, 09:04: PM

Expert Answer:

begin mathsize 16px style straight E equals 1 plus left parenthesis 1 plus straight x right parenthesis plus left parenthesis 1 plus straight x right parenthesis squared plus...... plus left parenthesis 1 plus straight x right parenthesis to the power of straight n
Here comma space straight a equals 1 comma space straight r equals left parenthesis 1 plus straight x right parenthesis
straight E equals fraction numerator open parentheses 1 plus straight x close parentheses to the power of straight n plus 1 end exponent minus 1 over denominator 1 plus straight x minus 1 end fraction
straight E equals 1 over straight x open square brackets open parentheses 1 plus straight x close parentheses to the power of straight n plus 1 end exponent minus 1 close square brackets
straight E equals straight x to the power of negative 1 end exponent open square brackets open parentheses 1 plus straight x close parentheses to the power of straight n plus 1 end exponent minus 1 close square brackets
If space you space see space the space above space equation space you space will space understand space this space coefficient space of space straight x to the power of straight k space
is space straight C presuperscript straight n plus 1 end presuperscript subscript straight k plus 1. end subscript
end style

Answered by Sneha shidid | 11th Dec, 2017, 02:14: PM