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Question: A series whose \(n^{th}\) term is \[\left( {n/x} \right) + y\], the sum of r terms will be 1) \[(r...

A series whose nthn^{th} term is (n/x)+y\left( {n/x} \right) + y, the sum of r terms will be

  1. (r(r+1)/2x)+ry(r(r + 1)/2x) + ry
  2. r(r−1)/2xr(r - 1)/2x
  3. (r(r−1)/2x)−ry(r(r - 1)/2x) - ry
  4. (r(r+1)/2y)+rx(r(r + 1)/2y) + rx
Explanation

Solution

the given simple problem can be solved easily as nthn^{th} term is given in the problem, we need to find the sum of rth term for that we need to find at least first 3 terms and rth term and we can obtain by giving the value of n as 1, 2, 3 and so on r. Then by adding all the terms she gets the required solution.

Complete step by step answer:
Now let us consider the given data
Since nthn^{th} term of the series is given as
an=1x+y{a_n} = \dfrac{1}{x} + y
Now let us find the first term by giving the value of n as 1 we get
a1=1x+y−−−(1){a_1} = \dfrac{1}{x} + y - - - \left( 1 \right)
To get the second term of the series given nis equal to 2 we get
a2=2x+y−−−(2){a_2} = \dfrac{2}{x} + y - - - \left( 2 \right)
Similarly put n as 3 we get the third term of the series and is given by
a3=3x+y−−−(3){a_3} = \dfrac{3}{x} + y - - - \left( 3 \right)
Now to get rth term replace n by r in the given series we get
ar=rx+y−−−(4){a_r} = \dfrac{r}{x} + y - - - \left( 4 \right)
Since we need to calculate the sum of r terms of the series, we can obtain it by adding equations 1, 2, 3, and 4 and is given by
Sum of r terms of the given series is =1x+y+2x+y+3x+y+−−−rx+y = \dfrac{1}{x} + y + \dfrac{2}{x} + y + \dfrac{3}{x} + y + - - - \dfrac{r}{x} + y
Since we have the term y r number of times so the above expression can be written as
Sum of r terms of the given series is =1x+2x+3x−−−rx+ry = \dfrac{1}{x} + \dfrac{2}{x} + \dfrac{3}{x} - - - \dfrac{r}{x} + ry
Taking (1x)\left( {\dfrac{1}{x}} \right) as a common factor we get
Sum of r terms of the given series is =(1x)(1+2+3+−−−+r)+ry = \left( {\dfrac{1}{x}} \right)\left( {1 + 2 + 3 + - - - + r} \right) + ry
We know that the sum of r term of a natural numbers 1+2+3+−−−+r1 + 2 + 3 + - - - + r by using this expression in the above step we get
Sum of r terms of the given series is =(1x)(r(r+1)2)+ry = \left( {\dfrac{1}{x}} \right)\left( {\dfrac{{r(r + 1)}}{2}} \right) + ry
Sum of r terms of the given series is =(r(r+1)2x)+ry = \left( {\dfrac{{r(r + 1)}}{{2x}}} \right) + ry

So, the correct answer is “Option 1”.

Note: A series can be highly generalized as the sum of all the terms in a sequence. However, there has to be a definite relationship between all the terms of the sequence.
In mathematics, a series is, roughly speaking, a description of the operation of adding infinitely many quantities, one after the other, to a given starting quantity. The study of series is a major part of calculus and its generalization, mathematical analysis.