sums | Eikonal Blog

2010.02.07

Simple sums

Filed under: mathematics — Tags: sums — sandokan65 @ 16:43

Definition:

$S_n^{(m)} :\equiv \sum_{k=1}^n k^m$
$A_n^{(m)} :\equiv \sum_{k=1}^n (-)^k k^m$

Several first sums:

$S_n^{(1)} = \frac12 n (n+1)$ ,
$S_n^{(2)} = \frac16 n (n+1) (2n+1)$ ,
$S_n^{(3)} = \frac14 n^2 (n+1)^2$ ,
$S_n^{(4)} = \frac1{30} n (n+1) (2n+1) (3n^2+3n-1)$ ,
$S_n^{(5)} = \frac1{12} n^2 (n+1)^2 (2n^2+2n-1)$ ,
…
$A_n^{(1)} = (-)^n \left[\frac{n+1}2\right]$ ,
$A_n^{(2)} = (-)^n \frac12 n(n+1)$ ,
$A_n^{(3)} = 4\left[\frac{n}2\right]^2 \left(\left[\frac{n}2\right]+1\right)^2 - \frac14 n^2(n+1)^2$ ,
$A_n^{(4)} = (-)^n \frac12 n(n^3+2n^2-1)$ ,
$A_n^{(5)} = -\frac14 [1+(-)^n(-2n^5 - 5n^4 + 5n^2 -1)]$ ,

Related sums:

$\sum_{k=0}^n (2k+1) = (n+1)^2$ ,
$\sum_{k=0}^n (2k+1)^2 = \frac13 (n+1)(2n+1)(2n+3)$ ,
$\sum_{k=0}^n (2k+1)^3 = (n+1)^2 (2n^2+4n+1)$ ,
$\sum_{k=0}^n (k+a)(k+b) = \frac16 n(n+1)[2n+1 + 3(a+b)]+nab$ ,
$\sum_{k=0}^n k(k+1) = \frac13 n(n+1)(n+2)$ ,
$\sum_{k=0}^n k(k+1)(k+2) = \frac14 n(n+1)(n+2)(n+3)$ ,
$\sum_{k=0}^n k(k+3)(k+6) = \frac14 n(n+1)(n+6)(n+7)$ ,
$\sum_{k=0}^n k(k+4)(k+8) = \frac14 n(n+1)(n+8)(n+9)$ ,
$\sum_{k=0}^n (-)^k (2k+1) = (-)^n (n+1)$ ,
$\sum_{k=0}^n (-)^k (2k+1)^2 = (-)^n 2(n+1)^2 - \frac12(1+(-)^n)$ .

General dependency:

$A_n^{(m)} = 2^{m+1} S_{[n/2]}^{(m)} - S_n^{(m)}$ .

More:

$\sum_{k=a}^b k^2 = a b n + \frac{n(n-1)(2n-1)}6$ = sum of squares of integer numbers, where $n= b-a + 1$ the number of terms in the sum.
$\sum_{k=a/2}^{b/2} (2k)^2 = a b n + \frac{2n(n-1)(2n-1)}3$ = sum of squares of even numbers.
$\sum_{k=(a-1)/2}^{(b-1)/2} (2k+1)^2 = a b n + \frac{2n(n-1)(2n-1)}3$ = sum of squares of odd numbers.
$\sum_{k=a}^b k^3 = \left[a b + \frac{n(n-1)}2 \right] \frac{n(a+b)}2$ = sum of cubes of integer numbers
$\sum_{k=a/2}^{b/2} (2k)^3 = \left[a b + 2 n(n-1) \right] \frac{n(a+b)}2$ = sum of cubes of even numbers.
$\sum_{k=(a-1)/2}^{(b-1)/2} (2k+1)^3 = \left[a b + 2 n(n-1) \right] \frac{n(a+b)}2$ = sum of cubes of odd numbers.
$\sum_{k=a}^b k^4 = a n \left[n b^2 + ab(n-1) \right] + \frac{n(n-1)(2n-1)}6 \cdot \frac{3n(n-1) -1}5$ = sum of fourth degrees of integer numbers.
$\sum_{k=a/2}^{b/2} (2k)^4 = a n \left[2 n b^2 - ab + ab(a-1) \right] + \frac{8 n(n-1)(2n-1)}6 \cdot \frac{3n(n-1) -1}5$ = sum of fourth degrees of even numbers.
$\sum_{k=(a-1)/2}^{(b-1)/2} (2k+1)^4 = a n \left[2 n b^2 - ab + ab(a-1) \right] + \frac{8 n(n-1)(2n-1)}6 \cdot \frac{3n(n-1) -1}5$ = sum of fourth degrees of odd numbers.
$\sum_{k=0}^{n-1} (a + d k)^3 = \left[a (a-d) + \frac{n(n-1) d}2 \right] \frac{n(a+b)}2$ = sum of cubes of numbers starting with $a$ , finishing with $b$ with every two neighboring members of sequence spread apart $d$ .

Sources:

T1989.02.14
“Summing same degrees of subsequent natural numbers” by N.A. Andreyev (Matematical Education #5, Moskva 1960; pp. 201-202) (in Russian)

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2010.01.05

Sums

Filed under: mathematics, number theory — Tags: series, sums — sandokan65 @ 15:39

$\sum_{k=1}^\infty \frac1{k^k} = 1.291,285,997,...$ (see Prudnikov 5.1.30.1)

$\sum_{k=1}^\infty \frac{(-)^k}{k} = -0.783,430,051,...$

$\sum_{k=1}^\infty \frac{k!}{k^k} = 1.879,853,86...$

$\sum_{k=1}^\infty \frac{(-)^k k!}{k^k} = -0.655,831,60...$

Reference: T1277

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2010.01.04

Binomial symbol

Filed under: mathematics — Tags: binomial symbol, math, sums — sandokan65 @ 19:56

Binomial theorem:
- “Source: The Functions of Mathematical Physics” by Harry Hochstadt
$\sum_{r=0}^m(-)^r \binom{n}{r} = (-)^n \binom{n-1}{m}$ for $0 \le m \le n-1$ .
$\sum_{r=k+1}^n (-)^{n-r} \binom{n}{r} = (-)^{n-k-1} \binom{n-1}{k}$ .

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