The value of $\sum\limits_{r = 0}^{1002} {{{( - 1)}^r}(a + rd)} $equals A.$a + 1003d$ B.$a + 1004d$ C.$a + 500d$ D.$a + 501d$
Answer
Verified
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’.
×
Sorry!, This page is not available for now to bookmark.