logo

Ask Questions, Get Answers

 
X
 Search
Want to ask us a question? Click here
Browse Questions
Ad
0 votes

Coeff. of $x^{50}$ in $(1+x)^{1000}$ + $x(1+x)^{999}$ + $x^2(1+x)^{998}$ + .... + $x^{1000}$ is

$\begin{array}{1 1} ^{1000} C_50 \\ ^{1001}C_{50} \\ ^{1001}C_{51} \\ ^{1000}C_{51} \end{array} $

Can you answer this question?
 
 

1 Answer

0 votes
$(1+x)^{1000}+x(1+x)^{999}+x^2(1+x)^{998}+......x^{1000}$ is a G.P.
$\therefore $ Sum of $1000 $ terms of this G.P. is
$(1+x)^{1000}\large\frac{\bigg[1-\bigg(\frac{x}{1+x}\bigg)^{1001}\bigg]}{1-\frac{x}{1+x}}$$=(1+x)^{1001}-x^{1001}$
$=^{1001}C_1x+^{1001}C_2x^2+..........$
$\therefore\:$ Coeff. of $x^{50} $ in the given G.P. is $^{1001}C_{50}$
answered Nov 23, 2013 by rvidyagovindarajan_1
 

Related questions

Ask Question
student study plans
x
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App
...