Certainly! Let's carefully evaluate the expression given:
[tex]\[
\sum_{j=1}^5 (d_{j+1} - d_j)
\][/tex]
This is a summation of the differences between successive terms in a sequence [tex]\(d\)[/tex]. We can write this summation explicitly as:
[tex]\[
(d_2 - d_1) + (d_3 - d_2) + (d_4 - d_3) + (d_5 - d_4) + (d_6 - d_5)
\][/tex]
Observe that this series is an example of a telescoping series, where most terms cancel out. Let's regroup and simplify it step by step:
1. Expanding the Summation:
[tex]\[
(d_2 - d_1) + (d_3 - d_2) + (d_4 - d_3) + (d_5 - d_4) + (d_6 - d_5)
\][/tex]
2. Rearranging to Highlight Cancellation:
It helps to see the structure of telescoping:
[tex]\[
d_2 - d_1 + d_3 - d_2 + d_4 - d_3 + d_5 - d_4 + d_6 - d_5
\][/tex]
3. Cancellation:
Notice that [tex]\(d_2\)[/tex] appears as [tex]\(+d_2\)[/tex] in the first term and [tex]\(-d_2\)[/tex] in the second term; similarly, [tex]\(d_3\)[/tex] as [tex]\(+d_3\)[/tex] and [tex]\(-d_3\)[/tex]; [tex]\(d_4\)[/tex] and [tex]\(d_5\)[/tex] likewise:
[tex]\[
\cancel{d_2} - d_1 + \cancel{d_3} - \cancel{d_2} + \cancel{d_4} - \cancel{d_3} + \cancel{d_5} - \cancel{d_4} + d_6 - \cancel{d_5}
\][/tex]
4. Simplified Result:
After canceling all intermediate terms, we are left with:
[tex]\[
d_6 - d_1
\][/tex]
Thus, the simplified form of the given summation is:
[tex]\[
d_6 - d_1
\][/tex]
This is the final result.
