sum of series to n term

3+18+57+132+255....................

We are given the series:

3, 18, 57, 132, 255, ...

We need to find the sum of the series up to the nth term.

Step 1: Analyze the pattern
First, observe the difference between consecutive terms:

18 - 3 = 15
57 - 18 = 39
132 - 57 = 75
255 - 132 = 123
The first differences are 15, 39, 75, 123, and they are increasing. Now, let’s find the second differences:

39 - 15 = 24
75 - 39 = 36
123 - 75 = 48
The second differences are 24, 36, 48, and they are increasing as well. Let’s now find the third differences:

36 - 24 = 12
48 - 36 = 12
The third differences are constant, meaning this series is a cubic sequence. So, we can represent the nth term using a cubic equation:

T(n) = an³ + bn² + cn + d

Step 2: Set up the system of equations
Using the first few terms of the sequence, we can write the following equations:

T(1) = 3 → a(1)³ + b(1)² + c(1) + d = 3 → a + b + c + d = 3
T(2) = 18 → a(2)³ + b(2)² + c(2) + d = 18 → 8a + 4b + 2c + d = 18
T(3) = 57 → a(3)³ + b(3)² + c(3) + d = 57 → 27a + 9b + 3c + d = 57
T(4) = 132 → a(4)³ + b(4)² + c(4) + d = 132 → 64a + 16b + 4c + d = 132
Step 3: Solve the system of equations
We now solve this system of linear equations:

a + b + c + d = 3
8a + 4b + 2c + d = 18
27a + 9b + 3c + d = 57
64a + 16b + 4c + d = 132
Subtract equation 1 from equation 2: (8a + 4b + 2c + d) - (a + b + c + d) = 18 - 3 7a + 3b + c = 15 → (Equation 5)

Subtract equation 2 from equation 3: (27a + 9b + 3c + d) - (8a + 4b + 2c + d) = 57 - 18 19a + 5b + c = 39 → (Equation 6)

Subtract equation 3 from equation 4: (64a + 16b + 4c + d) - (27a + 9b + 3c + d) = 132 - 57 37a + 7b + c = 75 → (Equation 7)

Now subtract equation 5 from equation 6: (19a + 5b + c) - (7a + 3b + c) = 39 - 15 12a + 2b = 24 6a + b = 12 → (Equation 8)

Now subtract equation 6 from equation 7: (37a + 7b + c) - (19a + 5b + c) = 75 - 39 18a + 2b = 36 9a + b = 18 → (Equation 9)

Now subtract equation 8 from equation 9: (9a + b) - (6a + b) = 18 - 12 3a = 6 a = 2

Substitute a = 2 into equation 8: 6(2) + b = 12 12 + b = 12 b = 0

Substitute a = 2 and b = 0 into equation 5: 7(2) + 3(0) + c = 15 14 + c = 15 c = 1

Finally, substitute a = 2, b = 0, and c = 1 into equation 1: 2 + 0 + 1 + d = 3 3 + d = 3 d = 0

Step 4: General formula for the nth term
Now we have the values for a, b, c, and d:

T(n) = 2n³ + n

Step 5: Sum of the first n terms
To find the sum of the first n terms, we use the formula for the sum of a cubic sequence:

S(n) = Σ (2n³ + n) from n = 1 to n

This can be written as: S(n) = 2Σn³ + Σn

The sum of the cubes of the first n natural numbers is: Σn³ = (n(n + 1) / 2)²

The sum of the first n natural numbers is: Σn = n(n + 1) / 2

Thus, the sum of the first n terms is: S(n) = 2(n(n + 1) / 2)² + n(n + 1) / 2

This gives the sum of the series up to the nth term.