| 初項a_0 公差b a_1 = a_0 +b a_2 = a_1 +b = a_0 + 2b a_3 = a_2 +b = a_0 + 3b a_4 = a_3 +b = a_0 + 4b … a_k = a_(k-1) +b = a_0+k b … |
| S_0 = a_0 S_1 = a_0+a_1 = 2a_0 +b S_2 = a_0+a_1+a_2 = 3a_0+b+2b = 3a_0+3b S_3 = a_0+a_1+a_2+a_3 = 4a_0 +b +2b +3b = 4a0+(1+2+3)b … S_n = (n+1)a_0 +(1+2+…+n)b = (n+1)a_0 +{n(n+1)/2}b {a_n} = {1,2,3,4,5,6,…,n,…｝ S_10 = 10*11/2 = 55 S_n = n(n+1)/2 {a_n} = {2n-1 | n = 1,2,3,…} S_n = 1+3+5+…+(2n-1) = n(2n-1) |
| {an} = {nm
| n = 1,2,3,…}, Sn=Σ[k=1,n] km
(m=1,2,3, …) |
| {an} = {n | n = 1,2,3,…} Sn = 1+2+3+…+n =(1/2) n(n+1) |
| {a_n} = {n2 | n = 1,2,3,…} S_n = 12+22+32+…+n2 =(1/6) n(n+1)(2n+1) |
| {a_n} = {n3 | n = 1,2,3,…} S_n = 13+23+33+…+n3 = (1/4)n2*(n+1)2 |
| {a_n} = {n4 | n = 1,2,3,…} S_n = 14+24+34+…+n4 = (1/30)n(n+1)(2n+1)(3n2 +3n-1) |
| {a_n} = {n5 | n = 1,2,3,…} S_n = 15+25+35+…+n5 = (1/12) n2(n+1)2(2n2 +2n-1) |
| {a_n} = {n6 | n = 1,2,3,…} S_n = 16+26+36+…+n6 = (1/42) n(n+1)(2n+1)(3n4+6n3n2 -3n+1) |
| {a_n} = {n7 | n = 1,2,3,…} S_n = 17+27+37+…+n7 = (1/24) n2(n+1)2(3n4+6n3-n2 -4n+2) |
| {a_n} = {n8 | n = 1,2,3,…} S_n = 18+28+38+…+n8 = (1/90) n(n+1)(2n+1)(5n6+15n5+5n4-15n3-n2 +9n-3) |
| {a_n} = {n9 | n = 1,2,3,…} S_n = 19+29+39+…+n9 = (1/20) n2(n+1)2(n2 +n-1)(2n4+4n3-n2 -3n+3) |
| {a_n} = {n10 | n = 1,2,3,…} S_n = 110+210+310+…+n10 = (1/66) n(n+1)(2n+1)(n2+n-1)(3n6+9n5+2n4-11n3+3n2 +10n -5) |
| an |
Sn=Σ[k=1,n} ak |
| 1 |
n |
| n |
(1/2) n(n+1) |
| n2 |
(1/6) n(n+1)(2n+1) |
| n3 |
(1/4) n2 (n+1)2 |
| n4 |
(1/30) n(n+1)(2n+1)(3n2+3n -1) |
| n5 |
(1/12) n2 (n+1)2 (2n2+2n
-1) |
| n6 |
(1/42) n(n+1)(2n+1)(3n4+6n3
-3n+1) |
| n7 |
(1/24) n2 (n+1)2 (3n4+6n3-n2
-4n+2) |
| n8 |
(1/90) n(n+1)(2n+1)(5n6+15n5+5n4
-15n3 -n2+9n -3) |
| wxMaximaによるΣ計算 |
|
| [演習] Maximaを使って 次式を計算せよ。 Σ[k=1,n] (k(k+1)(k+2) |
[解答] wxMaxima (%i 1) S:sum(k*(k+1)*(k+2),k,1,n),simpsum$ S1:factor(S); (%o2) (n*(n+1)*(n+2)*(n+3))/4 |
| ∑[k=1,n] G_k |
G_k = S_(k+1)-S_k |
| ∑[k=1,n] 1/((k(k+1)) |
∑(1/k)-1/(k+1) = 1-1/(n+1) |
| ∑[k=1,n] k*k! |
∑(k+1)!-k! = (n+1)! -1 |
| Sn=∑[k=1,n] k*xk-1 (x≠1) | Sn={1-(n+1)xn+nxn+1}/(1-x)2
|
| T_n = ∑[k=1,n] k*an =
n(n+1)(n+2)(n+3) T_(n+1) -T_n = (n+1)*a_(n+1) = 4(n+1)(n+2)(n+3) G_n = S_(n+1)-S_n = 4(n+2)(n+3) |
a_n = 4(n+1)(n+2) S_n =∑[k=1,n] a_n = 4∑[k=1,n] (n+1)(n+2) = 4n (n^2 +6n+11) |