Animated Solution for Mathematics - Sequence and Series: The sum of the first 20 terms of the series $5+11+19+29+41+\dots$ is

Imagine you are standing before a sequence of numbers: $5,11,19,29,41,\dots$. At first glance, it feels like a riddle. It is not an Arithmetic Progression because the difference between $11$ and $5$ is $6$, but the difference between $19$ and $11$ is $8$.

It is not a Geometric Progression either. This is where many students panic, but as an elite JEE aspirant, you should see this not as a problem, but as an invitation to investigate.

We start by calculating the first differences: $11-5=6$, $19-11=8$, $29-19=10$, and $41-29=12$. Look at that! The differences $6,8,10,12$ form a perfect Arithmetic Progression.

This is our 'Aha!' moment. When the first differences form an AP, the second differences—the difference of the differences—must be constant. Indeed, $8-6=2$, $10-8=2$, and $12-10=2$.

Because the second difference is constant, we know with mathematical certainty that the general term $T_n$ must be a quadratic polynomial of the form:

Now, how do we find that quadratic? We use the elegant 'Shift-and-Subtract' method. We write the sum $S_n$ in two rows, shifting the second row by one position.

When we subtract the second row from the first, the $S_n$ terms on the left cancel out to zero. On the right, we are left with the first term $5$, followed by the differences we just calculated, and finally, the negative of the $n^{th}$ term, $-T_n$.

This gives us the equation:

The bracketed part is an AP with $N=n-1$ terms, a first term $a=6$, and a common difference $d=2$. Using the sum formula $S=\frac{N}{2}[2a+(N-1)d]$, we substitute our values:

Simplifying this expression is a test of your algebraic discipline. Inside the bracket, we get $12+2n-4$, which simplifies to $2n+8$. Factoring out the $2$ cancels the denominator, leaving us with:

Expanding this, we arrive at the beautiful, clean general term:

Summation

We have conquered the hardest part. Now, the question asks for the sum of the first $20$ terms, $S_{20}={\sum }_{n=1}^{20}{T}_n$. Substituting our general term, we get:

Because the summation operator is linear, we can distribute it:

We apply the standard formulas: $\sum n^2=\frac{n(n+1)(2n+1)}{6}$, $\sum n=\frac{n(n+1)}{2}$, and $\sum 1=n$. For $n=20$:

1. The sum of squares is:

2. The sum of $n$ is $\frac{20\times 21}{2}=210$, and multiplying by $3$ gives $630$.

3. The sum of $1$ twenty times is $20$.

Adding these together: $2870+630+20=3520$. You have successfully navigated the complexity of the series to arrive at the final answer of 3520.