logo

Ask Questions, Get Answers

 
X
 Search
Want to ask us a question? Click here
Browse Questions
Ad
Home  >>  JEEMAIN and AIPMT  >>  Mathematics  >>  Class11  >>  Sequence and Series
0 votes

Given $1^2+2^2+3^2+....+2003^2$ = $(2003)(4007)(334)$ and $(1)(2003)+(2)(2002)+(3)(2001)+....+(2003)(1)$ = $(2003)(334)(x)$ then value of $x$ is

$(a)\;2001\qquad(b)\;2003\qquad(c)\;2005\qquad(d)\;2004$

Can you answer this question?
 
 

1 Answer

0 votes
Answer : (c) 2005
Explanation : $\;Let\;Q=1^2+2^2+....+2003^2$
$p=1(2003)+2(2002)+...+(2003)(1)$
$P+Q=(2004)\;[1+2+3+....2003]$
$(2003)(334)(4007+x)=\large\frac{(2004)(2003)(2004)}{2}$
$x=\large\frac{(2004)(1002)}{(334)}-4007$
$x=2005\;.$
answered Jan 23, 2014 by yamini.v 1 flag
 

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
...