1.試求下列級數1+2+2+3+3+3+4+4+4+4+...至第94項之和為_____
答:861
2.求值3·4＋4·5＋5·6+…＋52·53＝(A)44200 (B)49300 (C)49600 (D)50600
(3乘以4...)
答:c
3.試求1/(1*3)+1/(2*4)+1/(3*5)+1/(4*6)+1/(5*7)+...+1/(15*17)
答:375/544
駭客數學8-15
版主: thepiano
Re: 駭客數學8-15
重要公式
1 + 2 + 3 + ... + n = Σk = n(n + 1)/2 (k = 1 ~ n)
1^2 + 2^2 + 3^2 + ... + n^2 = Σk^2 = n(n + 1)(2n + 1)/6 (k = 1 ~ n)
第 1 題
1 + 2 + 3 + ... + 13 = 13(13 + 1)/2 = 91
所以第 92 ~ 94 項都是 14
原求值式 = 1^2 + 2^2 + 3^2 + 4^2 + ... + 13^2 + (14 * 3) = ......
第 2 題
原求值式 = Σk(k + 1) - (1 * 2 + 2 * 3) (k = 1 ~ 52)
= Σ(k^2 + k) - 8
= Σk^2 + Σk - 8
= ...
第 3 題
原求值式 = (1/2)[(1 - 1/3) + (1/3 - 1/5) + (1/5 - 1/7) + ... + (1/15 - 1/17)] + (1/2)[(1/2 - 1/4) + (1/4 - 1/6) + ... + (1/14 - 1/16)]
=(1/2)(1 - 1/17) + (1/2)(1/2 - 1/16)
= ...
1 + 2 + 3 + ... + n = Σk = n(n + 1)/2 (k = 1 ~ n)
1^2 + 2^2 + 3^2 + ... + n^2 = Σk^2 = n(n + 1)(2n + 1)/6 (k = 1 ~ n)
第 1 題
1 + 2 + 3 + ... + 13 = 13(13 + 1)/2 = 91
所以第 92 ~ 94 項都是 14
原求值式 = 1^2 + 2^2 + 3^2 + 4^2 + ... + 13^2 + (14 * 3) = ......
第 2 題
原求值式 = Σk(k + 1) - (1 * 2 + 2 * 3) (k = 1 ~ 52)
= Σ(k^2 + k) - 8
= Σk^2 + Σk - 8
= ...
第 3 題
原求值式 = (1/2)[(1 - 1/3) + (1/3 - 1/5) + (1/5 - 1/7) + ... + (1/15 - 1/17)] + (1/2)[(1/2 - 1/4) + (1/4 - 1/6) + ... + (1/14 - 1/16)]
=(1/2)(1 - 1/17) + (1/2)(1/2 - 1/16)
= ...