Find the sum:
(i) \(1+(-2)+(-5)+\cdots+(-236)\)
(ii) \(\big(4-\dfrac{1}{n}\big)+\big(4-\dfrac{2}{n}\big)+\cdots+\big(4-\dfrac{n}{n}\big)\)
(iii) \(\dfrac{a-b}{a+b}+\dfrac{3a-2b}{a+b}+\dfrac{5a-3b}{a+b}+\cdots\) to 11 terms
(i) \(-9400\)
(ii) \(\dfrac{7n-1}{2}\)
(iii) \(\dfrac{121a-66b}{a+b}\)
(i)
We are given the series: \(1, -2, -5, \dots, -236\).
This is an arithmetic progression (AP) because the difference between terms is constant.
First term (\(a\)) = 1.
Common difference (\(d\)) = \(-2 - 1 = -3\).
Last term (\(l\)) = -236.
Formula for last term: \(l = a + (n-1)d\).
Substitute: \(-236 = 1 + (n-1)(-3)\).
\(-236 - 1 = -3(n-1) \Rightarrow -237 = -3(n-1)\).
Divide both sides: \(n-1 = 79 \Rightarrow n = 80\).
Now, sum of \(n\) terms: \(S_n = \dfrac{n}{2}(a + l)\).
\(S_{80} = \dfrac{80}{2}(1 + (-236)) = 40(-235) = -9400\).
So, the sum is \(-9400\).
(ii)
The terms are: \(\big(4-\dfrac{1}{n}\big), \big(4-\dfrac{2}{n}\big), \dots, \big(4-\dfrac{n}{n}\big)\).
We can write the sum as:
\(S = \sum_{k=1}^{n} \left(4 - \dfrac{k}{n}\right)\).
Split into two parts:
\(S = \sum_{k=1}^{n} 4 - \sum_{k=1}^{n} \dfrac{k}{n}\).
First part: \(\sum_{k=1}^{n} 4 = 4n\).
Second part: \(\sum_{k=1}^{n} \dfrac{k}{n} = \dfrac{1}{n} \sum_{k=1}^{n} k\).
But \(\sum_{k=1}^{n} k = \dfrac{n(n+1)}{2}\).
So, second part = \(\dfrac{1}{n} \times \dfrac{n(n+1)}{2} = \dfrac{n+1}{2}\).
Therefore, \(S = 4n - \dfrac{n+1}{2}\).
Simplify: \(S = \dfrac{8n - (n+1)}{2} = \dfrac{7n - 1}{2}\).
So, the sum is \(\dfrac{7n-1}{2}\).
(iii)
We are asked to find the sum of 11 terms:
\(\dfrac{a-b}{a+b}, \dfrac{3a-2b}{a+b}, \dfrac{5a-3b}{a+b}, \dots\)
Notice that denominator in every term is the same: \((a+b)\).
So, we only need to add the numerators first.
Numerators: \((a-b), (3a-2b), (5a-3b), \dots\)
This forms an arithmetic progression (AP).
First numerator (\(a_1\)) = \(a-b\).
Second numerator = \(3a - 2b\).
So, common difference \(d = (3a - 2b) - (a - b) = 2a - b\).
Number of terms = 11.
Sum of numerators: \(S_{11} = \dfrac{11}{2}[2a_1 + (11-1)d]\).
Substitute: \(S_{11} = \dfrac{11}{2}[2(a-b) + 10(2a-b)]\).
\(= \dfrac{11}{2}[2a - 2b + 20a - 10b]\).
\(= \dfrac{11}{2}[22a - 12b]\).
\(= \dfrac{11}{2} \times 2(11a - 6b)\).
\(= 11(11a - 6b) = 121a - 66b\).
Now divide by denominator \((a+b)\).
So, required sum = \(\dfrac{121a - 66b}{a+b}\).