Sequences: arithmetic and geometric progressions

Pairing the first term with the last, the second with the second-to-last, and so on always gives the same pair-sum $u_1+u_n$, because moving one step forward from the start costs exactly $d$ and moving one step backward from the end gains back exactly $d$ — so the two changes cancel. This is the trick a schoolboy Gauss reportedly used to add $1+2+\cdots+100$ in seconds.

Write the sum forwards and then backwards, term by term: $S_n = u_n + u_{n-1} + \cdots + u_1$ is exactly the same sum, only listed in reverse order, so writing it directly beneath the forward sum $S_n=u_1+u_2+\cdots+u_n$ and adding the two equations column by column is legitimate.

Look at the $k$-th column of that addition: it is $u_k+u_{n+1-k}$. Since $u_k=u_1+(k-1)d$ and $u_{n+1-k}=u_1+(n-k)d$, adding them gives $u_k+u_{n+1-k}=2u_1+(n-1)d=u_1+u_n$. So every one of the $n$ columns produces the exact same value $u_1+u_n$, regardless of $k$ — this is precisely the cancellation described above.

Summing all $n$ columns therefore gives $2S_n = n(u_1+u_n)$, because the left side is $S_n+S_n$ and the right side is $n$ copies of the constant $u_1+u_n$.

Dividing both sides by $2$ gives $S_n=\dfrac{n(u_1+u_n)}{2}$. Substituting $u_n=u_1+(n-1)d$ into this expression and expanding gives the second form $S_n=nu_1+\dfrac{n(n-1)}{2}d$, which is useful when $u_n$ is not yet known. This argument never divided by anything that could be $0$ and never assumed $n$ is even (the pairing is purely algebraic column-addition, not a physical pairing-up of elements), so it holds for every $n\geq 1$.

Multiplying the whole sum $S_n$ by $q$ just shifts every term one position over, so subtracting $qS_n$ from $S_n$ makes almost every term cancel in a telescoping collapse, leaving only the very first and the very last (shifted) term. When $|q|<1$, repeatedly multiplying by $q$ shrinks a quantity toward $0$, so letting $n\to\infty$ the leftover term $u_1q^n$ simply vanishes and the finite formula turns into a fixed number.

Start from the definition $S_n = u_1+u_1q+u_1q^2+\cdots+u_1q^{n-1}$. Multiply both sides by $q$: $qS_n = u_1q+u_1q^2+\cdots+u_1q^{n-1}+u_1q^n$. Every term of $qS_n$ except the last, $u_1q^n$, already appears in $S_n$ shifted by one position.

Subtract: $S_n-qS_n$ cancels every shared term $u_1q,u_1q^2,\dots,u_1q^{n-1}$, leaving only the first term of $S_n$ (namely $u_1$) minus the last term of $qS_n$ (namely $u_1q^n$). This gives exactly $(1-q)S_n = u_1 - u_1 q^n$, i.e. $(1-q)S_n=u_1-u_1q^n$.

If $q\neq 1$, divide both sides by $1-q$ (legitimate since $1-q\neq 0$) to get $S_n = u_1\cdot\dfrac{1-q^n}{1-q}\quad (q\neq 1)$. If instead $q=1$, the telescoping identity reads $0\cdot S_n=0$, which carries no information, so this case must be handled directly from the definition: every term equals $u_1$, so $S_n=nu_1$.

Finally take $|q|<1$ and let $n\to\infty$ in the formula $S_n=u_1\cdot\dfrac{1-q^n}{1-q}$: since $|q|<1$ implies $q^n\to 0$ as $n$ grows, the numerator $1-q^n\to 1$, so $S_n\to \dfrac{u_1}{1-q}$. This limit is exactly $S = \dfrac{u_1}{1-q}\quad (|q|<1)$, the sum of the infinite geometric series; if instead $|q|\geq 1$ the term $q^n$ does not shrink to $0$ (it grows or stays constant in size), so the infinite sum does not exist as a finite number in that case.