# The value of $\sum\limits_{r = 0}^{1002} {{{( - 1)}^r}(a + rd)} $equals

A.$a + 1003d$

B.$a + 1004d$

C.$a + 500d$

D.$a + 501d$

Last updated date: 23rd Mar 2023

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Answer

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Hint: Here, we will solve the given question by summation of terms starting from r=0 to r=1002

Given,

$\sum\limits_{r = 0}^{1002} {{{( - 1)}^r}(a + rd)} \to (1)$

Now, let us expand the equation 1 by substituting the values of r from $0$ to 1002, we get

$ \Rightarrow {( - 1)^0}(a + (0)d) + {( - 1)^1}(a + d) + {( - 1)^2}(a + 2d) + ....... + {( - 1)^{1001}}(a + 1001d) + {( - 1)^{1002}}(a + 1002d)$Since, we know ${( - 1)^{odd}} = - 1$ and ${( - 1)^{even}} = 1$ , so the terms having r value as an odd number will have negative sign, whereas the terms having r value as an even number will have positive sign, now we can simplify the above equation as follows

$ \Rightarrow a - (a + d) + (a + 2d) + ....... - (a + 1001d) + (a + 1002d)$

Now, grouping the like terms, we get

$ \Rightarrow (a - a + a - a + ......$Upto $1003$ terms) -$d(1 - 2 + 3 - 4 + ... + 1001 - 1002) \to (2)$

As, we know 1-2+3-4…..upto n terms (‘n’ is even) =$ - \frac{n}{2}$. Here, we have n as 1002 which is even therefore,

$1 - 2 + 3 - 4 + .... + 1001 - 1002 = \frac{{ - 1002}}{2} = - 501 \to (3)$And

$a - a + a - a + ....upto 1003 terms = a[\because $Odd number of ‘a’ terms]$ \to (4)$

Substituting equations (3), (4) in equation (2), we get

$

\Rightarrow a - d( - 501) \\

\Rightarrow a + 501d \\

$

Therefore, $\sum\limits_{r = 0}^{1002} {{{( - 1)}^r}(a + rd)} = a + 501d$.

Hence, the correct option for the given question is ‘D’.

Note: Here, we have computed the value of a - a + a - a + ....upto 1003 terms as ‘a’. As out of 1003 terms, there will be 502 terms of ‘a’ and 501 terms ‘-a’. Hence, if we sum up the terms 501 terms of ‘a’ and 501 terms ‘-a’ gets cancelled and will be left with a single term ‘a’.

Given,

$\sum\limits_{r = 0}^{1002} {{{( - 1)}^r}(a + rd)} \to (1)$

Now, let us expand the equation 1 by substituting the values of r from $0$ to 1002, we get

$ \Rightarrow {( - 1)^0}(a + (0)d) + {( - 1)^1}(a + d) + {( - 1)^2}(a + 2d) + ....... + {( - 1)^{1001}}(a + 1001d) + {( - 1)^{1002}}(a + 1002d)$Since, we know ${( - 1)^{odd}} = - 1$ and ${( - 1)^{even}} = 1$ , so the terms having r value as an odd number will have negative sign, whereas the terms having r value as an even number will have positive sign, now we can simplify the above equation as follows

$ \Rightarrow a - (a + d) + (a + 2d) + ....... - (a + 1001d) + (a + 1002d)$

Now, grouping the like terms, we get

$ \Rightarrow (a - a + a - a + ......$Upto $1003$ terms) -$d(1 - 2 + 3 - 4 + ... + 1001 - 1002) \to (2)$

As, we know 1-2+3-4…..upto n terms (‘n’ is even) =$ - \frac{n}{2}$. Here, we have n as 1002 which is even therefore,

$1 - 2 + 3 - 4 + .... + 1001 - 1002 = \frac{{ - 1002}}{2} = - 501 \to (3)$And

$a - a + a - a + ....upto 1003 terms = a[\because $Odd number of ‘a’ terms]$ \to (4)$

Substituting equations (3), (4) in equation (2), we get

$

\Rightarrow a - d( - 501) \\

\Rightarrow a + 501d \\

$

Therefore, $\sum\limits_{r = 0}^{1002} {{{( - 1)}^r}(a + rd)} = a + 501d$.

Hence, the correct option for the given question is ‘D’.

Note: Here, we have computed the value of a - a + a - a + ....upto 1003 terms as ‘a’. As out of 1003 terms, there will be 502 terms of ‘a’ and 501 terms ‘-a’. Hence, if we sum up the terms 501 terms of ‘a’ and 501 terms ‘-a’ gets cancelled and will be left with a single term ‘a’.

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