Porisma

Worked solution

Expand sigma notation and count its terms

Write out each sum in full and state its number of terms: (i) ∑r=16r\sum_{r=1}^{6}r; (ii) ∑r=58r2\sum_{r=5}^{8}r^2; (iii) ∑k=04(2k−1)\sum_{k=0}^{4}(2k-1); (iv) ∑r=0∞(1/2)r\sum_{r=0}^{\infty}(1/2)^r.

CJC · Sequences and Series · Example 13

(i) Six consecutive integers

Answer

1+2+3+4+5+61+2+3+4+5+6; 66 terms.

Include both bounds

6−1+1=66-1+1=6

The source has already completed this row.

How did this part go?

(ii) Four squares

Answer

52+62+72+825^2+6^2+7^2+8^2; 44 terms.

Substitute every integer from five to eight

8−5+1=48-5+1=4

The upper bound is not the number of terms.

How did this part go?

(iii) Start at zero

Answer

[2(0)−1]+[2(1)−1]+[2(2)−1]+[2(3)−1]+[2(4)−1][2(0)-1]+[2(1)-1]+[2(2)-1]+[2(3)-1]+[2(4)-1]; 55 terms.

Preserve the first substituted index

4−0+1=54-0+1=5

The Teacher prints all five substitutions. Their simplified terms are −1,1,3,5,7.

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(iv) An infinite expansion

Answer

(12)0+(12)1+(12)2+⋯(\tfrac12)^0+(\tfrac12)^1+(\tfrac12)^2+\cdots; infinitely many terms.

Begin with the zeroth power

The expansion starts 1+1/2+1/4+… and never reaches a final index. The ask is to expand and count terms, not to evaluate the sum.

How did this part go?

Common mistakes

Use upper minus lower plus one only for finite integer bounds. An infinite number of terms does not by itself determine whether their sum converges.