Porisma

Worked solution

Choose a term formula and matching sigma bounds

Find an expression for the nth term and express each series in sigma notation: (a) 1+4+7+10+⋯+881+4+7+10+\cdots+88; (b) 1+3+5+7+⋯+991+3+5+7+\cdots+99; (c) −2+3−4+5−⋯+41-2+3-4+5-\cdots+41.

CJC · Sequences and Series · Example 14

(a) An arithmetic series ending at 88

Answer

un=3n−2u_n=3n-2 for 1≤n≤301\le n\le30; ∑r=130(3r−2)\sum_{r=1}^{30}(3r-2).

Form the general term

un=a+(n−1)d=1+(n−1)(3)=3n−2\begin{aligned}&u_n\\ & =a+(n-1)d\\ & =1+(n-1)(3)\\ & =3n-2\end{aligned}

The source identifies a=1,d=3.

Find the final index

88=1+(n−1)(3)⟹n=3088=1+(n-1)(3)\quad\Longrightarrow\quad n=30

The last term fixes the upper bound.

Use the same indexing in the sum

1+4+7+10+⋯+88=∑r=130(3r−2)\begin{aligned}&1+4+7+10+\cdots+88\\ & =\sum_{r=1}^{30}(3r-2)\end{aligned}

The first and final substitutions are 1 and 88.

How did this part go?

(b) Two indexings for odd terms

Answer

un=2n−1u_n=2n-1 for 1≤n≤501\le n\le50; ∑r=150(2r−1)=∑r=049(2r+1)\sum_{r=1}^{50}(2r-1)=\sum_{r=0}^{49}(2r+1).

Method 1: use the AP formula from position one

un=a+(n−1)d=1+(n−1)(2)=2n−1,99=1+(n−1)(2)⟹n=50,1+3+5+7+⋯+99=∑r=150(2r−1).\begin{aligned}u_n&=a+(n-1)d=1+(n-1)(2)=2n-1,\\99&=1+(n-1)(2)\quad\Longrightarrow\quad n=50,\\1+3+5+7+\cdots+99&=\sum_{r=1}^{50}(2r-1).\end{aligned}

The source gives a=1,d=2 and solves for fifty terms.

Method 2: start the counter at zero

vj=2j+1(0≤j≤49),1+3+5+7+⋯+99=∑r=049(2r+1)v_j=2j+1\quad(0\le j\le49),\qquad1+3+5+7+\cdots+99=\sum_{r=0}^{49}(2r+1)

The source writes 2n+1 for n≥0. This is a shifted indexing, not a second value for the same position-one u_n. Write v_j for that counter here; both sums retain the printed bounds.

How did this part go?

(c) Two indexings for alternating signs

Answer

un=(−1)n(n+1)u_n=(-1)^n(n+1) for 1≤n≤401\le n\le40; ∑r=140(−1)r(r+1)=∑r=241(−1)r+1r\sum_{r=1}^{40}(-1)^r(r+1)=\sum_{r=2}^{41}(-1)^{r+1}r.

Method 1: count positions from one

un=(−1)n(n+1),−2+3−4+5−⋯+41=∑r=140(−1)r(r+1)u_n=(-1)^n(n+1),\qquad-2+3-4+5-\cdots+41=\sum_{r=1}^{40}(-1)^r(r+1)

At position 1 the term is −2; at position 40 it is +41.

Method 2: use the magnitude as the counter

vj=(−1)j+1j(2≤j≤41),−2+3−4+5−⋯+41=∑r=241(−1)r+1rv_j=(-1)^{j+1}j\quad(2\le j\le41),\qquad-2+3-4+5-\cdots+41=\sum_{r=2}^{41}(-1)^{r+1}r

The source writes (−1)^(n+1)n from n=2. This shifted counter is denoted j here to distinguish it from the original position.

How did this part go?

Solution notes

The illustrated source guidance says sigma expressions are not unique: check the formula by substituting indices for its terms. A shifted counter needs shifted limits.