The sum of the coefficients of x 499 a n d x 500 in ( 1 + x ) 1000 + x ( 1 + x ) 999 + x 2 ( 1 + x ) 998 + ? + x 1000 is: [2026]

Question

The sum of the coefficients of $x^{499} and x^{500}$ in ${(1 + x)}^{1000} + x {(1 + x)}^{999} + x^{2} {(1 + x)}^{998} + \dots + x^{1000}$ is: [2026]

Smart Achievers · Accepted Answer

(1)

$S = {(1 + x)}^{1000} + x {(1 + x)}^{999} + x^{2} {(1 + x)}^{998} + \dots + x^{1000}$

$= {(1 + x)}^{1000} (\frac{1 - {(\frac{x}{1 + x})}^{1001}}{1 - \frac{x}{1 + x}})$

$= {(1 + x)}^{1001} - x^{1001}$

$Required sum = {}^{1001}{C_{499}} + {}^{1001}{C_{500}} = {}^{1002}{C_{500}}$

Solution

Q.
The sum of the coefficients of $x^{499} and x^{500}$ in ${(1 + x)}^{1000} + x {(1 + x)}^{999} + x^{2} {(1 + x)}^{998} + \dots + x^{1000}$ is: [2026]

1 ${}^{1002}C_{500}$

2 ${}^{1000}C_{501}$

3 ${}^{1001}C_{501}$

4 ${}^{1002}C_{501}$

Ans.
(1)

$S = {(1 + x)}^{1000} + x {(1 + x)}^{999} + x^{2} {(1 + x)}^{998} + \dots + x^{1000}$

$= {(1 + x)}^{1000} (\frac{1 - {(\frac{x}{1 + x})}^{1001}}{1 - \frac{x}{1 + x}})$

$= {(1 + x)}^{1001} - x^{1001}$

$Required sum = {}^{1001}{C_{499}} + {}^{1001}{C_{500}} = {}^{1002}{C_{500}}$

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Solution

Q. The sum of the coefficients of x499 and x500 in (1+x)1000+x(1+x)999+x2(1+x)998+⋯+x1000 is: [2026]

1 C5001002

2 C5011000

3 C5011001

4 C5011002

Ans. (1) S=(1+x)1000+x(1+x)999+x2(1+x)998+⋯+x1000 =(1+x)1000(1-(x1+x)10011-x1+x) =(1+x)1001-x1001 Required sum=C4991001 +C5001001=C5001002