the sum of n terms of the series
1, (1+2), (1+2+3)............is
A) (n/3)(n+1)(n-1) B)(n/6)(n+1)(n-1)
c) (n²+3n+20) D) none
try it yourselves, using the method I showed in your two other posts of this type
To find the sum of the series 1, (1+2), (1+2+3), and so on, we need to find the formula for the sum of the n terms of the series.
The sum of the first n natural numbers is given by the formula:
Sum = (n/2)(first term + last term)
In this series, the first term is always 1, and the last term for each term of the series is equal to the value of the term itself.
So, for the first term:
Sum1 = (1/2)(1 + 1) = 1
For the second term:
Sum2 = (2/2)(1 + (1+2)) = 4
For the third term:
Sum3 = (3/2)(1 + (1+2) + (1+2+3)) = 10
From these calculations, we can observe a pattern:
Sum1 = 1 = (1/2)(1)(2)
Sum2 = 4 = (2/2)(3)(4)
Sum3 = 10 = (3/2)(6)(7)
So, the sum of the n terms can be expressed as:
Sum(n) = (n/2)(n)(n+1)
Now, let's simplify this equation to check which option matches it.
Sum(n) = (n/2)(n)(n+1)
= (n/2)(n² + n)
= (n²/2) + (n/2)
Comparing this with the answer choices:
A) (n/3)(n+1)(n-1) - Not equal to the given equation.
B) (n/6)(n+1)(n-1) - Not equal to the given equation.
C) (n²+3n+20) - Not equal to the given equation.
D) None of the above.
Therefore, the correct answer is D) none.