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Question: The value of \(\sum\limits_{r = 0}^{1002} {{{( - 1)}^r}(a + rd)} \)equals A.\(a + 1003d\) B.\(a ...

The value of ∑r=01002(−1)r(a+rd)\sum\limits_{r = 0}^{1002} {{{( - 1)}^r}(a + rd)} equals
A.a+1003da + 1003d
B.a+1004da + 1004d
C.a+500da + 500d
D.a+501da + 501d

Explanation

Solution

Hint: Here, we will solve the given question by summation of terms starting from r=0 to r=1002
Given,
∑r=01002(−1)r(a+rd)→(1)\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 00 to 1002, we get
⇒(−1)0(a+(0)d)+(−1)1(a+d)+(−1)2(a+2d)+.......+(−1)1001(a+1001d)+(−1)1002(a+1002d)\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{( - 1)^{odd}} = - 1 and (−1)even=1{( - 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
⇒a−(a+d)+(a+2d)+.......−(a+1001d)+(a+1002d)\Rightarrow a - (a + d) + (a + 2d) + ....... - (a + 1001d) + (a + 1002d)
Now, grouping the like terms, we get
⇒(a−a+a−a+......\Rightarrow (a - a + a - a + ......Upto 10031003 terms) -d(1−2+3−4+...+1001−1002)→(2)d(1 - 2 + 3 - 4 + ... + 1001 - 1002) \to (2)
As, we know 1-2+3-4…..upto n terms (‘n’ is even) =−n2 - \frac{n}{2}. Here, we have n as 1002 which is even therefore,
1−2+3−4+....+1001−1002=−10022=−501→(3)1 - 2 + 3 - 4 + .... + 1001 - 1002 = \frac{{ - 1002}}{2} = - 501 \to (3)And
a−a+a−a+....upto1003terms=a[∵a - a + a - a + ....upto 1003 terms = a[\because Odd number of ‘a’ terms]→(4) \to (4)
Substituting equations (3), (4) in equation (2), we get
⇒a−d(−501) ⇒a+501d  \Rightarrow a - d( - 501) \\\ \Rightarrow a + 501d \\\
Therefore, ∑r=01002(−1)r(a+rd)=a+501d\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’.