(1) Find 1? + 2? + 3? +...+ 16? + 17? The value. Solution: 1? + 2? + 3? +...+ 16? + 17? = (6× 17? + 15× 17? + 10× 17? - 17)/30 = (85 19 142+ 12528 15+49 130- 17)/30 = 982 10738 + 3 + 4? + 5 + 6? +...+ 1 17 + 1 18? + 1 19 + 120? The value. Solution: 1+2? + 3 + 4? + 5 + 6? +...+ 1 17 + 1 18? + 1 19 + 120? = ( 1+3+5+...+ 1 19) + (2? +4? +6? +...+ 120? ) = ( 1+ 1 19)×[( 1 19- 1)÷2+ 1]/2 + 2? × ( 1? +2? +3? +...+60? ) = 1 20× 30+4× [60× (60+1)× (60× 2+1)/6] = 3600+295240 = 298840 (3) for/kloc. + 3? + 4 + 5? + 6? + 7 + 8? + 9? +...+ 4 15 + 4 16? + 4 17? + 4 18 + 4 19? + 420? The value. Solution: 1+2? + 3? + 4 + 5? + 6? + 7 + 8? + 9? +...+ 4 15 + 4 16? + 4 17? + 4 18 + 4 19? + 420? = ( 1+4+7+...+4 18) + (2? +5? +8? +...+4 19? ) + (3? +6? +9? +...+420? ) = ( 1+4 18)×[(4 18- 1)÷3+ 1]/2 + [(3×0+2)? +(3× 1+2)? +...+(3× 139+2)? ] + 3? × ( 1? +2? +3? +...+ 140? ) = 4 19×70 + [(3×0)? +(3× 1)? +...(3× 139)? ] + 2×2×(3×0+3× 1+...+3× 139) + 2? × 140 + 27×[ 140×( 140+ 1)/2]? = 29330 + (3? +6? +9? +...+4 17? ) + 4×3×( 1+2+...+ 139) + 560 + 2630256300 = 3? × ( 1? +2? +...+ 139? )+ 12×( 1+ 139)× 139/2+2630286 190 = 9×[ 139×( 139+ 1)×( 139×2+ 1)/6]+ 166...+n = n (n+ 1)/22. 1? + 2? + 3? +...+ n? = n(n+ 1)(2n+ 1)/6 ^ 3。 1? + 2? + 3? +...+ n? = [n(n+ 1)/2]? 4. 1? + 2? + 3? +...+ n? = (6n? + 15n? + 10n? -n)/30 . knowledge . Yahoo/question/question? Qid = 7011031901328 201-07-03 22: 31:46 Add: If we want to use more.
(2)
(3),a = 1/2。
b = 1/2
C = 0, so 1+2+3+...+n = n? /2+n/2 = n (n+1)/2.2011-07-03 22: 32: 43 Supplement: 2. Suppose 1? + 2? + 3? +...+ n? = Ann? + bn? ++d when n = 1, 1 = A+B+C+D-( 1) when n = 2,5 = 8a+4b+2c+d-1 4 = 27a+9b+3c+d-
(2)
(3)
(4),a = 1/3。
b = 1/2
c = 1/6
D = 0 2011-07-03 22: 40: 20 supplement: and so on. ...
Now expressed as ∑(k=4 18
K= 1) to ∑ is (3)
Wer = (1+4+27)+(∑ [(1+3k)+(2+3k) 2+(3+3k) 3) _ = (32)+[27 ∑ (k 3)+(27)
_ ∑ (k 3) = [(418) (419)/2] 2 _ ∑ (k 2) = [418 (419) (837).
∑( 1) =4 18 Subst。 The above item is converted into (1)
We have the answer = 2638546960.
1.327369 2.298403.660 can be calculated by Wolfram Alpha.
Reference: wolframalpha/