Answer: Options C and D are Correct
Given :
S= $\sum_{i=1}^{n} f(x_{i})$
A= $\sum_{i=1}^{a} f(x_{i})$
B=$\sum_{i=a}^{n} f(x_{i})$
Option A) It says S = A + B, which means $\sum_{i=1}^{a} f(x_{i})$ + $\sum_{i=a}^{n} f(x_{i})$, when summation will be done then the term f($x_{a}$) had been added two times. The first time in A and the second time in B. So Instead of S = A+B, it must be S= A+B- f($x_{a}$)
Option B) S=$\sum_{i=1}^{a-1} f(x_{i})$ + f($x_{a}$).+B
As we know B =$\sum_{i=a}^{n} f(x_{i})$, And in summation we are adding $f(x_a)$ so overcounting of f($x_{a}$) had been done in this case. So this option is Incorrect.
Option C)S =$\sum_{i=1}^{a-1} f(x_{i})$ +B
In this option, $\sum_{i=1}^{a-1} f(x_{i})$ + $\sum_{i=a}^{n} f(x_{i})$ had been done. So it can easily be observable that on the right-hand side of this option that each term is added once so we can write this as :
$\sum_{i=1}^{a-1} f(x_{i})$ + $\sum_{i=a}^{n} f(x_{i})$ =$\sum_{i=1}^{n} f(x_{i})$ = S
Actually here, The terms are broken down into two parts first from 1 to a-1 and second from a to n. So we are just adding both broken parts which give the result as from 1 to n.
Option c is correct.
Option D). S= A+B- f($x_{a}$)
Here while doing a summation of A and B the term f($x_{a}$) had been overcounted means counted two times but here, one time had been subtracted. So this option is also correct. As already explained in option A.
