Question

How do you solve for n in \[S = \dfrac{{n(n + 1)}}{2}\] ?

Answer 1

Hint: We have a simple equation with two variables or unknowns. After simplifying this we will get a polynomial of degree two. That is a quadratic equation. We can solve the obtained quadratic equation by factorization method. If the factorization method files we can use quadratic formula that is \[x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\] .

Complete step by step solution:
Given,
 \[S = \dfrac{{n(n + 1)}}{2}\] .
Multiply 2 on both sides of the equations,
 \[2S = n(n + 1)\]
 \[2S = {n^2} + n\]
 \[{n^2} + n - 2s = 0\]
Here we have a quadratic equation in the form \[a{n^2} + bn + c = 0\] .
The factorization method files because, the standard form of the factorization of quadratic equation is \[a{n^2} + {b_1}m + {b_2}m + c = 0\] , which does not satisfies the condition \[{b_1} \times {b_2} = a \times c\] and \[{b_1} + {b_2} = b\] .
So we use the quadratic formula,
 \[n = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\] , where \[a = 1,b = 1\] and \[c = - 2s\] .
Substituting we have,
 \[n = \dfrac{{ - 1 \pm \sqrt {{1^2} - 4\left( 1 \right)\left( { - 2s} \right)} }}{{2\left( 1 \right)}}\]
 \[ \Rightarrow n = \dfrac{{ - 1 \pm \sqrt {1 + 8s} }}{2}\] . This is the required answer.
So, the correct answer is “ \[ n = \dfrac{{ - 1 \pm \sqrt {1 + 8s} }}{2}\] ”.

Note: If we know the value of the ‘s’ we can find the value of ‘n’. We have a quadratic polynomial hence we have two roots, if we know the value of ‘s’ we will get two values for ‘n’. Also we know that the quadratic polynomial and Sirdar’s formula are same. we know that they have given the formula of sum of n natural number.