Sum an arithmetic progression

Derive the arithmetic-series formula by pairing

The source introduces the Gaussian method with a classroom story: a bored nine-year-old Gauss is told to add the integers from 1 to 100 as punishment, gives 5050 without writing, and surprises a teacher who checks by hand.

Treat these classroom details as the source’s story. The mathematical point is pattern recognition.

For historical context, Göttingen University records Gauss’s dates as 1777–1855 and his birthplace as Braunschweig; the lecture’s 1755 and Switzerland description are corrected here while the original wording remains in the native transcription.

Write the sum forwards and backwards:

$$\begin{aligned} S&=1+2+3+\cdots+98+99+100\quad\text{(1)},\\ S&=100+99+98+\cdots+3+2+1\quad\text{(2)}. \end{aligned}$$

Adding corresponding positions gives

$$2S=(1+100)+(2+99)+(3+98)+\cdots+(98+3)+(99+2)+(100+1).$$

There are 100 brackets, each 101, so $2S=100(101)=10100$ and $S=5050$. This adds two copies of the same finite sum; divide by two at the end.

For an AP with first term a and common difference d, the source applies the same idea to n terms:

$$\begin{aligned} S_n&=a+(a+d)+(a+2d)+\cdots\\ &\quad+[a+(n-3)d]+[a+(n-2)d]+[a+(n-1)d]\quad\text{(1)},\\ S_n&=[a+(n-1)d]+[a+(n-2)d]+[a+(n-3)d]+\cdots\\ &\quad+(a+2d)+(a+d)+a\quad\text{(2)}. \end{aligned}$$

Every corresponding pair has the same total. The printed pairing expansion is

$$\begin{aligned} 2S_n&=[a+a+(n-1)d]+[a+d+a+(n-2)d]+\cdots\\ &\quad+[a+(n-2)d+a+d]+[a+(n-1)d+a]\\ &=n[2a+(n-1)d],\\ S_n&=\frac n2[2a+(n-1)d]. \end{aligned}$$

The expanded lists illustrate the pattern for a sufficiently long AP; for any positive term count, pairing term j with term n+1−j gives the same proof without duplicating terms in an ellipsis. Since $u_n=a+(n-1)d$, the source also rewrites

$$\begin{aligned} S_n&=\frac n2[2a+(n-1)d]\\ &=\frac n2\{a+[a+(n-1)d]\}\\ &=\frac n2(a+u_n). \end{aligned}$$

The source’s final remark recommends $S_n=n(a+u_n)/2$ when the first and last terms are already known, rather than finding d unnecessarily. Both formulas express number of terms × average of the first and last terms.

Apply AP sums to totals and consecutive blocks

Example 19 uses $a=7,d=6,n=20$ to find the sum of $7,13,19,25,\ldots$:

$$S_{20}=\frac{20}{2}[2(7)+(20-1)(6)]=1280.$$

This is a sum of twenty terms, not the twentieth term found earlier. Example 20 supplies two consecutive blocks: the first seven terms total 28 and the next eight total 62. The second block is $S_{15}-S_7$, not $S_8$. With first term a and difference d,

$$\begin{aligned} S_7=\frac72(2a+6d)=28&\Longrightarrow7a+21d=28\\ &\Longrightarrow a+3d=4\quad\text{(1)},\\ S_{15}-S_7=62&\Longrightarrow\frac{15}{2}(2a+14d)-28=62\\ &\Longrightarrow15a+105d=90\\ &\Longrightarrow a+7d=6\quad\text{(2)}. \end{aligned}$$

The two equations give $a=5/2,d=1/2$, so the requested first three terms are $5/2,3,7/2$.

Self-Practice 6 instead supplies $u_9=43$ and $S_{15}=570$. Its printed answers are $a=3,d=5$ and least $n=31$ for $S_n>2265$. The worked exercise derives these from $a+8d=43$ and $a+7d=38$. The strict threshold matters: $S_{30}=2265$ is equal, whereas $S_{31}=2418$ exceeds it.

Choose a whole count in a capacity problem

In Example 21, Alice starts with 4800 sweets and fills bags with 25,30,35,…sweets. The source’s illustrated guidance identifies formulating and solving an inequality as the key. The total in n bags must satisfy

$$\frac n2[2(25)+(n-1)5]\le4800 \quad\Longleftrightarrow\quad n^2+9n-1920\le0.$$

For the analytical method, the quadratic roots are

$$n=\frac{-9\pm\sqrt{9^2-4(1)(-1920)}}{2(1)} =\frac{-9\pm\sqrt{7761}}2,$$

approximately −48.548 and 39.548. The source sketch is an upward-opening parabola crossing the horizontal axis at these roots. It is on or below the axis between them. This text describes the complete mathematical content of that sketch.

Restricting the exact interval to positive integer bag counts gives a maximum of 39.

The source’s truncated interval −48.54≤n≤39.54 has the same integer maximum but is not the exact real solution; keep exact roots until choosing the whole number.

The source’s graphical/calculator method enters Y1=(X/2)(50+5(X−1)). Its screen shows:

The written supporting table singles out 38→4465<4800, 39→4680<4800 and 40→4900>4800. Both methods therefore fill 39 bags and finish with

$$S_{39}=\frac{39}{2}[2(25)+38(5)]=4680,\qquad4800-4680=120$$

sweets left. The next bag would need 220 sweets, so 120 cannot fill it. Do not round the feasible count up.

Evaluate arithmetic series in sigma notation

The first n natural numbers form an AP with $a=1,d=1$ and last term n. Hence

$$1+2+3+\cdots+n=\sum_{r=1}^{n}r =\frac n2(a+u_n)=\frac n2(1+n).$$

Example 22 asks for both sigma notation and the sum of $1+4+7+10+\cdots+61$. With $a=1,d=3$, the last-term equation $1+(n-1)3=61$ gives $3n-2=61$ and n=21. The AP method gives

$$S_{21}=\frac{21}{2}[2(1)+(21-1)3]=651.$$

The second printed method uses the constant-factor and sum/difference rules:

$$\begin{aligned} 1+4+7+10+\cdots+61&=\sum_{r=1}^{21}(3r-2)\\ &=3\sum_{r=1}^{21}r-\sum_{r=1}^{21}2\\ &=3\left[\frac{21}{2}(1+21)\right]-2(21)=651. \end{aligned}$$

The source notes that AP concepts may be used when a question does not explicitly require sigma rules as its solving method. That does not remove a separate request to write the series in sigma notation.

Example 23 separates a linear term and a constant:

$$\begin{aligned} \sum_{k=1}^{n}(7k+5) &=7\sum_{k=1}^{n}k+\sum_{k=1}^{n}5\\ &=7(1+2+\cdots+n)+(5+5+\cdots+5)\\ &=7\left[\frac12n(n+1)\right]+5n\\ &=\frac12n(7n+17). \end{aligned}$$

The first group is an AP with first term 1 and difference 1; the second has n copies of 5. Both groups contain n terms.