Connect terms, partial sums and sigma notation

Check the first term after subtracting partial sums

Example 12 applies $u_n=S_n-S_{n-1}$ for $n\ge2$, then checks $u_1=S_1$ separately. With $S_n=2^n-1$, subtraction gives $u_n=2^{n-1}$; at n=1 this gives 1, which agrees with S1. The general ratio $u_{n+1}/u_n=2$ proves the terms form a GP.

For $S_n=n^2+3n$, the result is $u_n=2n+2$ for n≥2. Its n=1 value 4 agrees with S1, so the formula holds for every positive index. The general difference is $[2(n+1)+2]-(2n+2)=2$, proving an AP. The illustrated source callout asks you to show the common difference or ratio for general neighbouring terms, rather than simply presenting a familiar AP/GP-looking formula.

The extra constant in $S_n=n^2+3n+1$ cancels in the subtraction, but it changes the first term:

$$u_n=\begin{cases}5&n=1,\\2n+2&n\ge2.\end{cases}$$

The first three terms are 5,6,8. Their consecutive differences 1 and 2 disagree, as do their ratios 6/5 and 8/6. The full sequence is neither AP nor GP, even though its tail from the second term is arithmetic. The full worked exercise retains each printed expansion and first-term check.

Read every part of sigma notation

The Greek capital letter $\Sigma$, pronounced sigma, represents a sum. The source introduces it through $3^2+4^2+5^2+6^2+7^2+8^2+9^2$. Three pieces of information specify this sum: its general term, lower bound and upper bound. The source’s labelled notation diagram has four labels:

This table gives a text equivalent of all the diagram’s labels. For integer bounds $m\le n$,

$$\sum_{r=m}^{n}u_r=u_m+u_{m+1}+\cdots+u_{n-1}+u_n.$$

In the opening example,

$$\sum_{r=3}^{r=9}r^2=\sum_{r=3}^{9}r^2=3^2+4^2+5^2+6^2+7^2+8^2+9^2.$$

Read this as “the sum of all terms of the form r squared, where r takes each integer value from 3 to 9 inclusive”. Omitting the repeated r= above sigma does not change the meaning. The index need not be r: $\sum_{k=1}^{5}k^2=\sum_{r=1}^{5}r^2$, and

$$\sum_{r=1}^{n}(2r+1)=\sum_{k=1}^{n}(2k+1)=\sum_{i=1}^{n}(2i+1).$$

Only the bound counter and its matching occurrences are renamed. The source’s Equivalence callout notes that different sigma forms can make simplification and manipulation easier.

Change the counter together with the bounds

Sigma expressions are not unique. The source gives

$$\sum_{r=1}^{10}r=\sum_{r=0}^{9}(r+1)$$

and, for a positive integer n,

$$\sum_{r=1}^{2n}r(r+1) =\sum_{r=0}^{2n-1}(r+1)(r+2) =\sum_{r=-2n-1}^{-2}r(r+1).$$

The source ends these equalities with ellipses, not a supplied evaluated answer. The last form runs through the same products in reverse order: set $r=-k-1$ to obtain $r(r+1)=k(k+1)$. This substitution is an authored explanation of the printed equivalence. Integer counters can be zero or negative when the formula is defined; their bounds need not start at one.

Typical term patterns are $2r$ for even integers, $2r+1$ or $2r-1$ for odd integers, and $(-1)^r$, $(-1)^{r+1}$ or $(-1)^{r-1}$ for alternating signs. Choose the exponent to match the first sign at the chosen starting index. Check both endpoint substitutions whenever you change the formula or limits.

Expand, count and construct sums

With finite integer bounds, including both endpoints gives

$$\text{number of terms}=n-m+1\qquad\text{in }\sum_{r=m}^{n}u_r.$$

Example 13 expands six integers from 1 to 6, four squares from 5 to 8, five terms of $2k-1$ from 0 to 4, and infinitely many powers $(1/2)^r$ from 0 onwards. The last expansion starts with $(1/2)^0=1$. Its infinite number of terms does not by itself tell you the sum’s value.

Example 14 constructs the formula and its bounds together. The AP $1+4+7+10+\cdots+88$ has $a=1,d=3$, so $u_n=1+(n-1)3=3n-2$. Solving $88=1+(n-1)3$ gives 30 terms and $\sum_{r=1}^{30}(3r-2)$.

For the odd terms through 99, the two printed methods give

$$\sum_{r=1}^{50}(2r-1)=\sum_{r=0}^{49}(2r+1).$$

The position-one nth-term formula is $u_n=2n-1$; a counter starting at 0 instead uses $v_j=2j+1$, $0\le j\le49$. The source calls both formulas u_n; the different letters here make the change of indexing explicit rather than giving two values to one position.

For $-2+3-4+5-\cdots+41$, the two printed methods give

$$\sum_{r=1}^{40}(-1)^r(r+1)=\sum_{r=2}^{41}(-1)^{r+1}r.$$

The first indexing has $u_n=(-1)^n(n+1)$, $1\le n\le40$; the shifted magnitude counter has $v_j=(-1)^{j+1}j$, $2\le j\le41$. Both begin at −2 and finish at +41. The illustrated source guidance stresses checking an expression by substituting its indices for each term.

Self-Practice 4 retains both printed alternatives for each series:

$$1^4+2^4+\cdots+n^4+(n+1)^4 =\sum_{r=1}^{n+1}r^4=\sum_{r=0}^{n}(r+1)^4,$$ $$-1+2-3+4-5+6 =\sum_{r=1}^{6}(-1)^r r =\sum_{r=2}^{7}(-1)^{r-1}(r-1).$$

The working panels keep all original methods and term-count results. Do not replace a requested expansion with only its numerical total.

Evaluate a finite sum with the supplied calculator

The source illustrates $\sum_{r=2}^{8}r^4(r-1)$ using its calculator:

1. Press MATH, scroll to option 0 (summation). Alternatively press ALPHA WINDOW, then choose option 2.
2. Enter lower limit 2, upper limit 8 and formula X^4(X−1). X is the calculator’s counter corresponding to the written r.
3. Press Enter. The displayed result is 53004.

The original menu, template/input and result screens are retained in the illustration archive. The worked exercise also gives an independent seven-term arithmetic check.

Self-Practice 5 prints three results: the fifteen squares $1^2+2^2+\cdots+15^2$ sum to 1240; $\sum_{r=1}^{20}(3-2r)=-360$; and $\sum_{r=0}^{15}2^r=65535$. The last has sixteen terms because it includes index 0. Its working shows cancellation after doubling. Summation laws and AP/GP sum formulas provide further ways to evaluate larger sums.

Find a least index from partial sums

Example 15 asks for the least positive integer $n$ with $H_n=\sum_{r=1}^{n}1/r>4$. The source enters a summation in Y1, using X for its upper bound and a separate R as the counter:

$$Y_1(X)=\sum_{R=1}^{X}\frac1R.$$

Its displayed table has these values, retaining the screen’s precision:

The illustrated guidance requires a supporting table when using this calculator method. The printed written table gives $H_{30}\approx3.995<4$, $H_{31}\approx4.027>4$ and $H_{32}\approx4.059>4$. The last rounded value is a source error: the unrounded $H_{32}\approx4.058495195437$ rounds to 4.058 at three decimal places. Avoid rounding an already rounded display a second time.

Each additional reciprocal is positive, so the partial sums strictly increase. Since $H_{30}<4<H_{31}$, the least index is 31. This monotonicity argument explains why checking the neighbouring values establishes the first crossing; it supplements the printed calculator evidence.