1. Identify the pattern of the sequence

Given sequence: $1,\;3,\;11,\;25,\;45,\;71,\ldots$

Let us examine successive differences: $3-1=2,$ $11-3=8,$ $25-11=14,$ $45-25=20,$ $71-45=26.$

So the first differences are: $2,\;8,\;14,\;20,\;26,\ldots$

These form an arithmetic progression with common difference $6$.

2. Write recurrence for the $n$th term

If $a_n$ is the sequence, then $a_n-a_{n-1}=2,8,14,20,26,\ldots$ This is an AP whose $(n-1)$th term is: $d_n=2+6(n-2)=6n-10 \quad (n\ge 2).$

Thus, $a_n=a_{n-1}+(6n-10), \quad a_1=1.$

3. Find explicit formula for $a_n$

We have $a_n=1+\sum_{k=2}^n (6k-10).$

Now, $\sum_{k=2}^n (6k-10)=6\sum_{k=2}^n k-10(n-1).$

Using $\sum_{k=1}^n k=\frac{n(n+1)}{2},$ we get $\sum_{k=2}^n k=\frac{n(n+1)}{2}-1.$

So, $a_n=1+6\left(\frac{n(n+1)}{2}-1\right)-10(n-1).$

Simplify: $a_n=1+3n(n+1)-6-10n+10$ $=3n^2+3n-10n+5$ $=3n^2-7n+5.$

Check:

• $n=1$: $3-7+5=1$
• $n=2$: $12-14+5=3$
• $n=3$: $27-21+5=11$

Correct.

4. Find the sum of first 20 terms

We need $S_{20}=\sum_{n=1}^{20} (3n^2-7n+5).$

So, $S_{20}=3\sum_{n=1}^{20} n^2 -7\sum_{n=1}^{20} n +5\sum_{n=1}^{20} 1.$

Now use standard formulas: $\sum_{n=1}^{20} n=\frac{20\cdot21}{2}=210,$ $\sum_{n=1}^{20} n^2=\frac{20\cdot21\cdot41}{6}=2870,$ $\sum_{n=1}^{20} 1=20.$

Therefore, $S_{20}=3(2870)-7(210)+5(20).$

Compute: $3(2870)=8610,$ $7(210)=1470,$ $5(20)=100.$

Hence, $S_{20}=8610-1470+100=7240.$

5. Match with the options

$S_{20}=7240$

So the correct option is: A: 7240

6. Comparison with stored correct answer

Stored correct answer: A

Derived answer: A

They agree.