Evaluate $\displaystyle\sum_{k=1}^{11} (2 + 3^k)$.

Evaluate Σk=1¹¹ (2 + 3k).

Exercise 8.2 · Q11

\displaystyle\sum_{k=1}^{11} (2 + 3^k)

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Concept understanding — Sum Of Series

What Does "Sum of a Series" Even Mean?

Imagine you're standing at point 0 and you take a step of 1 metre forward. Then a step of half a metre. Then a quarter metre. Then an eighth. And you keep going, each step half the size of the previous one.

After 1 step: you're at 1 metre.

After 2 steps: at 1.5 metres.

After 3 steps: at 1.75 metres.

After 4 steps: at 1.875 metres.

After 5 steps: at 1.9375 metres.

You notice something: you're getting closer and closer to 2 metres, but you never quite reach it. If you could take infinitely many steps, would you ever get to exactly 2 metres? This is the heart of what a series is — adding up infinitely many numbers and asking: does this sum settle down to a finite value?

Note

A series is just the sum of the terms of a sequence. If the sequence is $a_1, a_2, a_3, \dots$ , then the series is $a_1 + a_2 + a_3 + \dots$ .

The Precise Definition

Let’s formalise this. Suppose we have an infinite sequence of numbers:

a_1, a_2, a_3, a_4, \dots

We want to make sense of the infinite sum:

a_1 + a_2 + a_3 + a_4 + \dots

We can't just "add infinitely many things" directly — that's not a finite operation. So mathematicians do something clever: they look at partial sums.

Define:

S_1 = a_1

S_2 = a_1 + a_2

S_3 = a_1 + a_2 + a_3

\vdots

S_n = a_1 + a_2 + \dots + a_n

$S_n$ is called the $n$ th partial sum — it's the sum of the first $n$ terms.

Now, the series is said to converge (or have a sum) if the sequence of partial sums $S_1, S_2, S_3, \dots$ approaches some finite number $S$ as $n$ gets larger and larger. In that case, we write:

\sum_{k=1}^{\infty} a_k = S

If the partial sums don't settle down to a finite number — they either grow without bound or oscillate forever — the series diverges and has no finite sum.

\sum_{k=1}^{\infty} a_k = \lim_{n \to \infty} S_n \quad \text{where} \quad S_n = \sum_{k=1}^{n} a_k

The Walking Example, Formalised

Our sequence of steps: $1, \frac12, \frac14, \frac18, \dots$

The partial sums:

S_1 = 1

S_2 = 1 + \frac12 = 1.5

S_3 = 1 + \frac12 + \frac14 = 1.75

S_4 = 1 + \frac12 + \frac14 + \frac18 = 1.875

You can prove (and we will later) that:

S_n = 2 - \frac{1}{2^{n-1}}

As $n \to \infty$ , $\frac{1}{2^{n-1}} \to 0$ , so $S_n \to 2$ . Therefore:

\sum_{k=1}^{\infty} \frac{1}{2^{k-1}} = 2

The infinite sum equals exactly 2 — even though you never "reach" it after any finite number of steps, the limit of the process is 2.

Two Classic Examples to Build Intuition

1. The Harmonic Series (Diverges)

1 + \frac12 + \frac13 + \frac14 + \frac15 + \dots

This one is tricky. The terms get smaller and smaller, but the partial sums grow without bound — just very slowly. $S_{100} \approx 5.18$ , $S_{1000} \approx 7.48$ , $S_{1000000} \approx 14.39$ . It never stops growing. This series diverges.