Finding the nth Term Using Sum Formula

The problem asks us to find the general formula for the nth term (\(a_n\)) of a series, given the formula for the sum of its first n terms (\(S_n\)).

We are given:

• The sum of the first n terms: \(S_n = n(2n+1)\)

Calculating the nth Term (\(a_n\))

The nth term of a series (\(a_n\)) can be found by taking the difference between the sum of the first n terms (\(S_n\)) and the sum of the first (n-1) terms (\(S_{n-1}\)). The formula is:

\(a_n = S_n - S_{n-1}\)

Let's calculate \(S_{n-1}\) using the given formula for \(S_n\). We replace 'n' with '(n-1)':

\(S_{n-1} = (n-1)(2(n-1)+1)\)

Now, let's simplify the expression for \(S_{n-1}\):

1. Distribute the 2 inside the parenthesis: \(S_{n-1} = (n-1)(2n - 2 + 1)\)
2. Combine the constants inside the parenthesis: \(S_{n-1} = (n-1)(2n - 1)\)
3. Expand the expression by multiplying the terms (using the FOIL method: First, Outer, Inner, Last): \(S_{n-1} = n(2n) + n(-1) - 1(2n) - 1(-1)\) \(S_{n-1} = 2n^2 - n - 2n + 1\)
4. Combine like terms: \(S_{n-1} = 2n^2 - 3n + 1\)

Now we have the formulas for \(S_n\) and \(S_{n-1}\). Let's substitute them into the formula for \(a_n\):

\(a_n = S_n - S_{n-1}\) \(a_n = [n(2n+1)] - [2n^2 - 3n + 1]\)

First, expand the expression for \(S_n\):

\(S_n = 2n^2 + n\)

Now substitute the expanded \(S_n\) back into the equation for \(a_n\):

\(a_n = (2n^2 + n) - (2n^2 - 3n + 1)\)

Distribute the negative sign to the terms in the second parenthesis:

\(a_n = 2n^2 + n - 2n^2 + 3n - 1\)

Group and combine like terms:

\(a_n = (2n^2 - 2n^2) + (n + 3n) - 1\) \(a_n = 0 + 4n - 1\) \(a_n = 4n - 1\)

So, the nth term of the series is \(4n - 1\). This matches option 1.

Verification (Optional)

Let's check for the first few terms:

• For n=1: \(S_1 = 1(2(1)+1) = 1(3) = 3\). The first term \(a_1\) should be \(4(1)-1 = 3\). This matches.
• For n=2: \(S_2 = 2(2(2)+1) = 2(5) = 10\). The second term \(a_2\) should be \(4(2)-1 = 7\). \(S_2 = a_1 + a_2 = 3 + 7 = 10\). This matches.
• For n=3: \(S_3 = 3(2(3)+1) = 3(7) = 21\). The third term \(a_3\) should be \(4(3)-1 = 11\). \(S_3 = a_1 + a_2 + a_3 = 3 + 7 + 11 = 21\). This matches.

The calculation is confirmed.