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Mathematics · Ch 11 — Sequences and Series

Properties of Summation and Standard Results

11.7.2

Properties of Summation and Standard Results

Properties of Summation: (i) ∑r=1nktr=k∑r=1ntr\sum_{r=1}^n kt_r=k\sum_{r=1}^n t_r for a nonzero constant kk; (ii) ∑r=1n(ar+br)=∑r=1nar+∑r=1nbr\sum_{r=1}^n(a_r+b_r)=\sum_{r=1}^na_r+\sum_{r=1}^nb_r; (iii) ∑r=1n1=n\sum_{r=1}^n 1=n; (iv) ∑r=1nk=kn\sum_{r=1}^{n} k = kn for a nonzero constant kk (a restatement of (i) and (iii) together).

Result 1: the sum of the first nn natural numbers is ∑r=1nr=n(n+1)2\sum_{r=1}^{n}r=\dfrac{n(n+1)}2.

Result 2: the sum of the squares of the first nn natural numbers is ∑r=1nr2=n(n+1)(2n+1)6\sum_{r=1}^{n}r^2=\dfrac{n(n+1)(2n+1)}6.

Result 3: the sum of the cubes of the first nn natural numbers is ∑r=1nr3=[n(n+1)2]2\sum_{r=1}^{n}r^3=\left[\dfrac{n(n+1)}2\right]^2 (these three results can be formally proved using Mathematical Induction, covered later in the book).

Worked Example 1: evaluate ∑r=1n(8r−7)\sum_{r=1}^{n}(8r-7). =8∑r−7∑1=8⋅n(n+1)2−7n=4n2+4n−7n=4n2−3n=8\sum r-7\sum1=8\cdot\dfrac{n(n+1)}2-7n=4n^2+4n-7n=4n^2-3n.

Worked Example 2: find 32+42+52+⋯+2923^2+4^2+5^2+\cdots+29^2. =∑r=129r2−∑r=12r2=29⋅30⋅596−2⋅3⋅56=(29×5×59)−5=5(29×59−1)=5(1710)=8550=\sum_{r=1}^{29}r^2-\sum_{r=1}^{2}r^2=\dfrac{29\cdot30\cdot59}6-\dfrac{2\cdot3\cdot5}6=(29\times5\times59)-5=5(29\times59-1)=5(1710)=8550.

Worked Example 3: find 1002−992+982−972+⋯+22−12100^2-99^2+98^2-97^2+\cdots+2^2-1^2. Grouping the even-indexed and odd-indexed squares separately: =∑r=150(2r)2−∑r=150(2r−1)2=∑r=150(4r−1)=4⋅50⋅512−50=5100−50=5050=\sum_{r=1}^{50}(2r)^2-\sum_{r=1}^{50}(2r-1)^2=\sum_{r=1}^{50}(4r-1)=4\cdot\dfrac{50\cdot51}2-50=5100-50=5050. …