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Question: Summation of series (method of differences), when we use \[\sum\limits_{{}}^{{}}{f\left( r \right)-f...

Summation of series (method of differences), when we use ∑f(r)−f(r+1)\sum\limits_{{}}^{{}}{f\left( r \right)-f\left( r+1 \right)}, how can we know it is f(1)−f(n+1)f\left( 1 \right)-f\left( n+1 \right)but not f(n)−f(1+1)f\left( n \right)-f\left( 1+1 \right)? Also, how can we know it is f(n+1)−f(1)f\left( n+1 \right)-f\left( 1 \right), but not f(1+1)−f(n)f\left( 1+1 \right)-f\left( n \right) when we use ∑f(r+1)−f(r)\sum\limits_{{}}^{{}}{f\left( r+1 \right)-f\left( r \right)}?

Explanation

Solution

In order to solve the given question, we need to apply the summation of ∑f(r)−f(r+1)\sum\limits_{{}}^{{}}{f\left( r \right)-f\left( r+1 \right)}and simplified it to get the required answer. Here we substitute r = 1, 2, 3, 4, ………., n-1, n. Similarly, again we need to apply the summation of ∑f(r+1)−f(r)\sum\limits_{{}}^{{}}{f\left( r+1 \right)-f\left( r \right)} and simplifies it further, we will get the required answer.

Complete step by step solution:
We know that,
⇒S=∑r=1n(f(r)−f(r+1))\Rightarrow S=\sum\limits_{r=1}^{n}{\left( f\left( r \right)-f\left( r+1 \right) \right)}
Now, we substitute r = 1, 2, 3, 4, ………., n-1, n
We get,
⇒S=∑r=1n(f(r)−f(r+1))=(f(1)−f(2))+(f(2)−f(3))+(f(3)−f(4))+........+(f(n−1)−f(n))+(f(n)−f(n+1))\Rightarrow S=\sum\limits_{r=1}^{n}{\left( f\left( r \right)-f\left( r+1 \right) \right)}=\left( f\left( 1 \right)-f\left( 2 \right) \right)+\left( f\left( 2 \right)-f\left( 3 \right) \right)+\left( f\left( 3 \right)-f\left( 4 \right) \right)+........+\left( f\left( n-1 \right)-f\left( n \right) \right)+\left( f\left( n \right)-f\left( n+1 \right) \right)Simplifying the above, we get
All the terms cancels out except f(1) and f(n+1).
⇒S=∑r=1n(f(r)−f(r+1))=f(1)−f(n+1)\Rightarrow S=\sum\limits_{r=1}^{n}{\left( f\left( r \right)-f\left( r+1 \right) \right)}=f\left( 1 \right)-f\left( n+1 \right)
Therefore, Summation of series (method of differences), when we use ∑f(r)−f(r+1)\sum\limits_{{}}^{{}}{f\left( r \right)-f\left( r+1 \right)} is always equal tof(1)−f(n+1)f\left( 1 \right)-f\left( n+1 \right).
Similarly,
We know that,
⇒S=∑r=1n(f(r+1)−f(r))\Rightarrow S=\sum\limits_{r=1}^{n}{\left( f\left( r+1 \right)-f\left( r \right) \right)}
Now, we substitute r = 1, 2, 3, 4, ………., n-1, n
We get,
⇒S=∑r=1n(f(r+1)−f(r))=(f(2)−f(1))+(f(3)−f(2))+(f(4)−f(3))+........+(f(n)−f(n−1))+(f(n+1)−f(n))\Rightarrow S=\sum\limits_{r=1}^{n}{\left( f\left( r+1 \right)-f\left( r \right) \right)}=\left( f\left( 2 \right)-f\left( 1 \right) \right)+\left( f\left( 3 \right)-f\left( 2 \right) \right)+\left( f\left( 4 \right)-f\left( 3 \right) \right)+........+\left( f\left( n \right)-f\left( n-1 \right) \right)+\left( f\left( n+1 \right)-f\left( n \right) \right)Simplifying the above, we get
All the terms cancels out each other except f(1) and f(n+1).
⇒S=∑r=1n(f(r+1)−f(r))=−f(1)+f(n+1)\Rightarrow S=\sum\limits_{r=1}^{n}{\left( f\left( r+1 \right)-f\left( r \right) \right)}=-f\left( 1 \right)+f\left( n+1 \right)
We can also rewrite it as,
⇒S=∑r=1n(f(r+1)−f(r))=f(n+1)−f(1)\Rightarrow S=\sum\limits_{r=1}^{n}{\left( f\left( r+1 \right)-f\left( r \right) \right)}=f\left( n+1 \right)-f\left( 1 \right)

Therefore, Summation of series (method of differences), when we use ∑f(r+1)−f(r)\sum\limits_{{}}^{{}}{f\left( r+1 \right)-f\left( r \right)} is always equal tof(n+1)−f(1)f\left( n+1 \right)-f\left( 1 \right).

Note: Whenever we get these types of questions, students should always remember the concept of summation of series and a way we solve the summation of series. Students should carefully write all the numbers very explicitly as there are always high chances of mistakes in the calculation part. Always observe the negative and positive sign of a number very keenly to avoid making errors.