Algebra theory - sequences

Sum of Arithmetic progression:

Classic “reverse-and-add” proof for the sum of an arithmetic progression

Let

$$S_n \;=\; a + \bigl(a+d\bigr) + \bigl(a+2d\bigr) + \dots + \bigl[a+(n-1)d\bigr],$$

where

• $a$ is the first term,

• $d$ is the common difference,

• $n$ is the number of terms.

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1. Write the same sum in reverse order

$$S_n \;=\; \bigl[a+(n-1)d\bigr] + \bigl[a+(n-2)d\bigr] + \dots + a.$$

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2. Add the two expressions term-by-term

$$\begin{aligned} S_n &{}= a + (a+d) + (a+2d) + \dots + \bigl[a+(n-1)d\bigr] \\ S_n &{}= \bigl[a+(n-1)d\bigr] + \bigl[a+(n-2)d\bigr] + \dots + a \\ \hline 2S_n &{}= \bigl[\,a + a + (n-1)d\,\bigr] \;+\; \bigl[\,a + a + (n-2)d\,\bigr] \;+\; \dots \;+\; \bigl[\,a + a + 0\cdot d\,\bigr]. \end{aligned}$$

Notice every bracket on the right is the same:

$$a + a + (n-1)d \;=\; 2a + (n-1)d.$$

Because there are $n$ brackets, we have

$$2S_n \;=\; n\bigl[\,2a + (n-1)d\,\bigr].$$

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3. Solve for $S_n$

$$\boxed{\,S_n \;=\; \dfrac{n}{2}\,\bigl[\,2a + (n-1)d\,\bigr]\,}.$$

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4. Alternate form using the last term

Let the last term be $l = a + (n-1)d$.
Then

$$S_n \;=\; \frac{n}{2}\,(a + l),$$

the well-known “average of first and last, times number of terms.”

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Why it works

Reversing the list pairs each smallest term with the matching largest term, the next-smallest with the next-largest, and so on. Every pair sums to the same constant $a + l$; there are $n$ such pairs, so the double-sum $2S_n$ is $n(a+l)$. Dividing by 2 gives the desired formula.

This elegant trick—often credited to a young Gauss—works for any arithmetic progression, no matter the starting value $a$, common difference $d$, or length $n$.

Sum of Geometric progression:

“Multiply-and-subtract” proof for the finite geometric series

Let

$$S_n \;=\; a + ar + ar^{2} + \dots + ar^{\,n-1},$$

where

• $a$ is the first term,

• $r$ is the common ratio,

• $n$ is the number of terms.

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1. Multiply the entire sum by the common ratio $r$

$$rS_n \;=\; ar + ar^{2} + ar^{3} + \dots + ar^{\,n}.$$

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2. Subtract the second equation from the first

$$\begin{aligned} S_n - rS_n &= \bigl(a + ar + ar^{2} + \dots + ar^{\,n-1}\bigr) \;-\; \bigl(ar + ar^{2} + \dots + ar^{\,n}\bigr) \\ &= a - ar^{\,n}. \end{aligned}$$

Every intermediate term cancels: $ar$ cancels, $ar^{2}$ cancels, and so on, leaving only the first term $a$ and the “extra” term $ar^{\,n}$ from the shifted series.

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3. Solve for $S_n$

$$(1 - r)S_n \;=\; a\bigl(1 - r^{\,n}\bigr) \quad\Longrightarrow\quad \boxed{\,S_n \;=\; \dfrac{a\bigl(1 - r^{\,n}\bigr)}{1 - r}\,}\quad(r \neq 1).$$

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4. Special case $r = 1$

If $r = 1$, every term is $a$, so

$$S_n \;=\; a + a + \dots + a \;=\; na.$$

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5. Infinite-series extension ( $|r| < 1$ )

Taking the limit as $n \to \infty$ with $|r| < 1$:

$$S_\infty \;=\; \lim_{n\to\infty} S_n \;=\; \frac{a}{1 - r},$$

because $r^{\,n} \to 0$.

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Why it works

Multiplying by $r$ shifts every term one place to the right. Subtracting aligns the two sums so that all interior terms cancel pairwise, isolating just the first term of the original series and the “overflow” term at the end. The resulting simple linear equation in $S_n$ yields the closed-form formula.

Example 1:

1) Find sum of $\sum_{x=1}^{n} x^2 \cdot 2^x$

Solution:
Whenever you see a series which is "partially" G.P., you can use the same trick we used to prove G.P. sum formula, i.e. multiply by common ratio and subtract.
Here we will have to do it twice.

$$S_n = 1^2 \cdot 2 + 2^2 \cdot 2^2 + 3^2 \cdot 2^3 + \cdots + n^2 \cdot 2^n$$ $$2S_n=1^2\cdot 2^2+2^2\cdot 2^3+\cdots +(n-1{)}^2\cdot 2^n+n^2\cdot 2^{n+1}$$

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$$- S_n = 1 \cdot 2 + 3 \cdot 2^2 + 5 \cdot 2^3 + \cdots + (2n - 1) \cdot 2^n - n^2 \cdot 2^{n+1}$$ $$- 2S_n = 1 \cdot 2^2 + 3 \cdot 2^3 + \cdots + (2n - 3) \cdot 2^n + (2n - 1) \cdot 2^{n+1} - n^2 \cdot 2^{n+2}$$

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$$S_n = 1 \cdot 2 + 2 \cdot 2^2 + 2 \cdot 2^3 + \cdots + 2 \cdot 2^n$$ $$S_n = 2 + 8 \left( \frac{1 - 2^{n-1}}{1 - 2} \right) - n^2 \cdot 2^{n+1} - (2n - 1) \cdot 2^{n+1} + n^2 \cdot 2^{n+2}$$

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$$S_n = 2 + 8 \left(2^{n-1} - 1\right) - 2^{n+1} \left(n^2 - 2n + 1\right) + n^2 \cdot 2^{n+2}$$ $$S_n = 2^{n+2} - 6 - (n-1)^2 \cdot 2^{n+1} + n^2 \cdot 2^{n+2}$$

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Example 2:

$$\sum_{r=1}^{n} (r+1) \cdot 3^r$$

Solution:
It's quite simple.
Split it into 2 sums.
One of them is a simple G.P.
For the other we will multiply by common ratio and subtract once.

For

$$S_n \;=\;\sum_{r=1}^{n}(r+1)\,3^{\,r},$$

split the summand into two simpler series:

$$S_n=\sum_{r=1}^{n}r\,3^{\,r}\;+\;\sum_{r=1}^{n}3^{\,r}.$$

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1. The geometric part

$$\sum_{r=1}^{n}3^{\,r}=3\;\frac{3^{\,n}-1}{3-1}= \frac{3}{2}\bigl(3^{\,n}-1\bigr).$$

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2. The “$r\,3^{r}$” part

A standard closed form is

$$\sum_{r=1}^{n} r\,a^{r}= \frac{a\bigl[1-(n+1)a^{n}+n\,a^{n+1}\bigr]}{(1-a)^{2}} \qquad(a\neq1).$$

With $a=3$:

$$\sum_{r=1}^{n} r\,3^{\,r}= \frac{3\bigl[1-(n+1)3^{\,n}+n\,3^{\,n+1}\bigr]}{(1-3)^{2}} =\frac{3}{4}\Bigl[1+(2n-1)3^{\,n}\Bigr].$$

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3. Combine

$$\begin{aligned} S_n &=\frac{3}{4}\Bigl[1+(2n-1)3^{\,n}\Bigr] +\frac{3}{2}\bigl(3^{\,n}-1\bigr) \\ &= \frac{3}{4}\Bigl[(2n+1)3^{\,n}-1\Bigr]. \end{aligned}$$

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Final closed form

$$\boxed{\displaystyle S_n=\frac{3}{4}\Bigl[(2n+1)3^{\,n}-1\Bigr]}$$

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Quick verification

$n$	Direct sum $\sum_{r=1}^{n}(r+1)3^{r}$	Formula $\dfrac{3}{4}\bigl[(2n+1)3^{n}-1\bigr]$	Match?
1	$(1+1)3^{1}=2\cdot3=6$	$\dfrac{3}{4}\bigl[(3)3-1\bigr]=\dfrac{3}{4}(9-1)=6$	✅
2	$(1+1)3^{1}+(2+1)3^{2}=6+27=33$	$\dfrac{3}{4}\bigl[(5)9-1\bigr]=\dfrac{3}{4}(45-1)=33$	✅

Both checks confirm the closed-form formula is correct.