Question

If ${1^2} + {2^2} + {3^2} + ......... + {2003^2} = \left( {2003} \right)\left( {4007} \right)\left( {334} \right)$ and $1\left( {2003} \right) + 2\left( {2002} \right) + 3\left( {2001} \right) + ......... + 2003\left( 1 \right) = \left( {2003} \right)\left( {334} \right)\left( x \right)$ , then x equals
A.2005
B.2004
C.2003
D.2001

Answer 1

We will find the value of ‘x’ by simplifying the equation containing ‘x’. for that we will simplify the left-hand side expression using some summation properties like
$\sum\limits_{r = 1}^n {A - B} = \sum\limits_{r = 1}^n A - \sum\limits_{r = 1}^n B$
$\sum\limits_{r = 1}^n {kA(r)} = k\sum\limits_{r = 1}^n {A(r)}$
$\sum\limits_{r = 1}^n r = \dfrac{{n\left( {n + 1} \right)}}{2}$
Then we will compare it with the expression of the right-hand side to get the required value of x.

Complete step-by-step answer:
Given data: ${1^2} + {2^2} + {3^2} + ......... + {2003^2} = \left( {2003} \right)\left( {4007} \right)\left( {334} \right)$
$1\left( {2003} \right) + 2\left( {2002} \right) + 3\left( {2001} \right) + ......... + 2003\left( 1 \right) = \left( {2003} \right)\left( {334} \right)\left( x \right)$
Solving for the $1\left( {2003} \right) + 2\left( {2002} \right) + 3\left( {2001} \right) + ......... + 2003\left( 1 \right) = \left( {2003} \right)\left( {334} \right)\left( x \right)$
Taking the left-hand side of the equation
$= 1\left( {2003} \right) + 2\left( {2002} \right) + 3\left( {2001} \right) + ......... + 2003\left( 1 \right)$
From the above left-hand side expression, we can find the general term of the expression let say ${T_r}$
$\therefore {T_r} = r\left( {2003 - r + 1} \right)A + S$
And since the left-hand side is the sum of all the terms we can say that
Therefore the left-hand side will be $\sum\limits_{r = 1}^{2003} {r\left( {2003 - r + 1} \right)}$
On simplifying we get,
$\Rightarrow \sum\limits_{r = 1}^{2003} {r\left( {2003 - r + 1} \right)} = \sum\limits_{r = 1}^{2003} {r\left( {2004 - r} \right)}$
On simplifying the brackets we get,
$= \sum\limits_{r = 1}^{2003} {\left( {2004r - {r^2}} \right)}$
Now we know that $\sum\limits_{r = 1}^n {A - B} = \sum\limits_{r = 1}^n A - \sum\limits_{r = 1}^n B A + S$
$= \sum\limits_{r = 1}^{2003} {2004r} - \sum\limits_{r = 1}^{2003} {{r^2}}$
We know that $\sum\limits_{r = 1}^n {kA(r)} = k\sum\limits_{r = 1}^n {A(r)}$, and writing the second by expanding as the sum of its terms
$= 2004\sum\limits_{r = 1}^{2003} r - \left( {{1^2} + {2^2} + {3^2}......... + {{2003}^2}} \right)$
Now it is well known that
$\sum\limits_{r = 1}^n r = \dfrac{{n\left( {n + 1} \right)}}{2}$, and it is given that ${1^2} + {2^2} + {3^2} + ......... + {2003^2} = \left( {2003} \right)\left( {4007} \right)\left( {334} \right)$
Therefore, using these
$= 2004\left[ {\dfrac{{2003\left( {2003 + 1} \right)}}{2}} \right] - \left( {2003} \right)\left( {4007} \right)\left( {334} \right)$
On division and simplification we get,
$= \left( {2004} \right)\left( {2003} \right)\left( {1002} \right) - \left( {2003} \right)\left( {4007} \right)\left( {334} \right)$
Writing 2004 as a product of its factors i.e. $2004 = 334 \times 6$
$= \left( {334 \times 6} \right)\left( {2003} \right)\left( {1002} \right) - \left( {2003} \right)\left( {4007} \right)\left( {334} \right)$
Now taking $\left( {2003} \right)\left( {334} \right)$ as common from both the terms
$= \left( {2003} \right)\left( {334} \right)\left[ {\left( 6 \right)\left( {1002} \right) - \left( {4007} \right)} \right]$
On simplifying the square brackets we get,
$= \left( {2003} \right)\left( {334} \right)\left[ {6012 - 4007} \right]$
On simplifying, we get
$= \left( {2003} \right)\left( {334} \right)\left( {2005} \right)$
Now comparing the left-hand side with the right-hand side, i.e.
$\Rightarrow \left( {2003} \right)\left( {334} \right)\left( {2005} \right) = \left( {2003} \right)\left( {334} \right)\left( x \right)$
We conclude that, $x = 2005$
Hence, Option (A) is correct.

Note: In the above solution, we have a term as $\sum\limits_{r = 1}^{2003} {{r^2}}$, most of the students its value using the formula i.e. $\sum\limits_{r = 1}^n {{r^2}} = \dfrac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6}$
Therefore using this formula we will get,
$\sum\limits_{r = 1}^{2003} {{r^2}} = \dfrac{{2003\left( {2004} \right)\left( {4007} \right)}}{6}$
Dividing the numerator and the denominator by 6
$[\sum\limits_{r = 1}^{2003} {{r^2}} = 334\left( {2003} \right)\left( {4007} \right)]$, which is equivalent to the value given in the question.
So, this method is not incorrect but since its value is given in the question we just have to mold it in that form, so try to imply all the data given to us in place of finding the same given value.