Summation of sequence

(a) There are n terms in the n th group, so there are totally n(n+1)/2 numbers,
and the last term is n(n+1)/2. Using sum of AS = n[T(1)+T(n)]/2, so,
the sum of these numbers is :
[n(n+1)/2][1 + n(n+1)/2]/2
= n(n + 1)(n^2 + n + 2)/8

(b) Same as before, using sum of GS = T(1)*(r^n - 1)/(r - 1), so,
the sum of these numbers is :
1*{2^[n(n+1)/2] - 1}/(2 - 1)
= 2^[n(n+1)/2] - 1

(a)

consider the last term in each group,
group 1, the last term = 1 = 1(1 + 1)/2
group 2, the last term = 3 = 1 + 2 = 2(2 + 1)/2
group 3, the last term = 6 = 1 + 2 + 3 = 3(3 + 1)/2
group 4, the last term = 10 = 1 + 2 + 3 + 4 = 4(4 + 1)/2
so the last term of group n can be written as the form of n(n + 1)/2

thus, the sum to n groups
= 1 + 2 + 3 + ... + n(n + 1)/2
= [n(n + 1)/2][n(n + 1)/2 + 1]/2
= [(n^2 + n)/2][(n^2 + n + 2)/2]/2
= (n^2 + n)(n^2 + n + 2)/8

(b)

consider the last term in each group,
group 1, the last term = 1 = 2^0 = 2^[1(1 + 1)/2 - 1]
group 2, the last term = 4 = 2^2 = 2^[2(2 + 1)/2 - 1]
group 3, the last term = 32 = 2^5 = 2^[3(3 + 1)/2 - 1]
group 4, the last term = 512 = 2^9 = 2^[4(4 + 1)/2 - 1]
so the last term of group n can be written as the form of 2^[n(n + 1)/2 - 1]

thus, the sum to n groups
= 1 + 2 + 4 + ... + 2^[n(n + 1)/2 - 1]
= 2^0 + 2^1 + 2^2 + ... + 2^[n(n + 1)/2 - 1]
= 2^0 + 2^1 + 2^2 + ... + 2^[(n^2 + n - 2)/2]
= 2^0{2^[(n^2 + n - 2)/2] - 1}/(2 - 1)
= 2^[(n^2 + n - 2)/2]

2012-09-21 08:32:16 補充：
= 2^[(n^2 + n - 2)/2] - 1

a 1+2+3+4+.... +n
用等差數列總和公式
=項數(首項+末項)/2
=n(n+1)/2 //

b 1+2+4+...
=2^0 + 2^1 +2^2 +... +2^(n-1)
用等比數列總和公式
=首項(公比n次方 -1) / (公比-1)
=1(2^n - 1) / (2-1)
=2^n - 1 //