Finding the Sum of the Series 1 + 9 + 24 + 46 + 75 + ......

The given series is \(S_n = 1 + 9 + 24 + 46 + 75 + \dots\) up to n terms. Let the terms of the series be denoted by \(T_1, T_2, T_3, \dots, T_n\).

The terms are \(T_1 = 1\), \(T_2 = 9\), \(T_3 = 24\), \(T_4 = 46\), \(T_5 = 75\), and so on.

Determining the General Term of the Series

To find the sum of n terms of this series, we first need to find the general term \(T_n\). Let's look at the differences between consecutive terms:

• Difference between 2nd and 1st term: \(T_2 - T_1 = 9 - 1 = 8\)
• Difference between 3rd and 2nd term: \(T_3 - T_2 = 24 - 9 = 15\)
• Difference between 4th and 3rd term: \(T_4 - T_3 = 46 - 24 = 22\)
• Difference between 5th and 4th term: \(T_5 - T_4 = 75 - 46 = 29\)

The sequence of these first differences is 8, 15, 22, 29, ...

Now, let's find the differences between consecutive terms of this new sequence (the second differences):

• Difference between 2nd and 1st first difference: \(15 - 8 = 7\)
• Difference between 3rd and 2nd first difference: \(22 - 15 = 7\)
• Difference between 4th and 3rd first difference: \(29 - 22 = 7\)

Since the second differences are constant (equal to 7), the general term \(T_n\) of the original series is a quadratic expression in n. We can write \(T_n\) in the form \(An^2 + Bn + C\), where A, B, and C are constants we need to determine.

Finding the Coefficients A, B, and C

We can use the first few terms of the series to form equations:

• For n=1: \(T_1 = A(1)^2 + B(1) + C = A + B + C = 1\)
• For n=2: \(T_2 = A(2)^2 + B(2) + C = 4A + 2B + C = 9\)
• For n=3: \(T_3 = A(3)^2 + B(3) + C = 9A + 3B + C = 24\)

Now, we solve this system of linear equations:

Subtract the first equation from the second:

\((4A + 2B + C) - (A + B + C) = 9 - 1\)

\(3A + B = 8\) (Equation 1)

Subtract the second equation from the third:

\((9A + 3B + C) - (4A + 2B + C) = 24 - 9\)

\(5A + B = 15\) (Equation 2)

Subtract Equation 1 from Equation 2:

\((5A + B) - (3A + B) = 15 - 8\)

\(2A = 7\)

\(A = \frac{7}{2}\)

Substitute the value of A into Equation 1:

\(3\left(\frac{7}{2}\right) + B = 8\)

\(\frac{21}{2} + B = 8\)

\(B = 8 - \frac{21}{2} = \frac{16 - 21}{2} = -\frac{5}{2}\)

Substitute the values of A and B into the first original equation \(A + B + C = 1\):

\(\frac{7}{2} + \left(-\frac{5}{2}\right) + C = 1\)

\(\frac{2}{2} + C = 1\)

\(1 + C = 1\)

\(C = 0\)

So, the general term is \(T_n = \frac{7}{2}n^2 - \frac{5}{2}n = \frac{n(7n - 5)}{2}\).

Calculating the Sum of n Terms

The sum of n terms, \(S_n\), is the sum of the general term \(T_k\) from \(k=1\) to \(n\):

\(S_n = \sum_{k=1}^{n} T_k = \sum_{k=1}^{n} \left(\frac{7}{2}k^2 - \frac{5}{2}k\right)\)

We can split the sum into two parts:

\(S_n = \frac{7}{2} \sum_{k=1}^{n} k^2 - \frac{5}{2} \sum_{k=1}^{n} k\)

Using the standard formulas for the sum of the first n integers (\(\sum_{k=1}^{n} k = \frac{n(n+1)}{2}\)) and the sum of the first n squares (\(\sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}\)), we substitute these into the expression for \(S_n\):

\(S_n = \frac{7}{2} \left(\frac{n(n+1)(2n+1)}{6}\right) - \frac{5}{2} \left(\frac{n(n+1)}{2}\right)\)

\(S_n = \frac{7n(n+1)(2n+1)}{12} - \frac{5n(n+1)}{4}\)

To combine these terms, find a common denominator, which is 12. Multiply the second term by \(\frac{3}{3}\):

\(S_n = \frac{7n(n+1)(2n+1)}{12} - \frac{5n(n+1) \times 3}{4 \times 3}\)

\(S_n = \frac{7n(n+1)(2n+1)}{12} - \frac{15n(n+1)}{12}\)

Now, factor out the common terms \(\frac{n(n+1)}{12}\):

\(S_n = \frac{n(n+1)}{12} \left(7(2n+1) - 15\right)\)

\(S_n = \frac{n(n+1)}{12} \left(14n + 7 - 15\right)\)

\(S_n = \frac{n(n+1)}{12} \left(14n - 8\right)\)

Factor out 2 from the term in the parenthesis:

\(S_n = \frac{n(n+1)}{12} \times 2(7n - 4)\)

\(S_n = \frac{n(n+1)(7n - 4)}{6}\)

This formula gives the sum of the first n terms of the series.

Comparing with Options

Let's compare our derived sum formula with the given options:

• Option 1: \(\frac{n(n + 1)(5n - 4)}{6}\)
• Option 2: \(\frac{n(n + 1)(3n - 4)}{6}\)
• Option 3: \(\frac{n(n - 1)(7n + 4)}{6}\)
• Option 4: \(\frac{n(n + 1)(7n - 4)}{6}\)

Our derived formula is \(\frac{n(n + 1)(7n - 4)}{6}\), which matches Option 4.