Given that the sequence {an} satisfies: A1 = A2 − 2A + 2, an + 1 = an + 2 (n − a) + 1, n ∈ n *, if and only if n = 3, an is the minimum, then the value range of real number a is () A. （-1，3）B. (52，3)C. (52，72)D. （2，4）

Given that the sequence {an} satisfies: A1 = A2 − 2A + 2, an + 1 = an + 2 (n − a) + 1, n ∈ n *, if and only if n = 3, an is the minimum, then the value range of real number a is () A. （-1，3）B. (52，3)C. (52，72)D. （2，4）

From an + 1 = an + 2 (n-a) + 1: A2 = a1 + 2 (1-A) + 1 & nbsp; & nbsp; & nbsp; A3 = A2 + 2 (2-A) + 1 & nbsp; & nbsp; & nbsp; A4 = A3 + 2 (3-A) + 1 &The sum of an = an-1 + 2 (n-1-a) + 1 is: an = a1 + 2 [1 + 2 + 3 + +(n-1) - (n-1) a] + n-1 = a1 + 2 (n − 1) N2 − 2 (n − 1) a + n − 1 because A1 = A2 − 2A + 2, so an = A2 − 2A + 2 + N2 − n − 2An + 2A + n − 1 = n2-2an + A2-1, Let f (n) = an = N2 − 2An + A2 − 1, the opening of the function is upward, and the equation of symmetry axis is n = − 2A2 = a, because n ∈ n *, so when 52 ＜ a ＜ 72, f (n) = an is the minimum, so select C

Given that the sequence {an} satisfies A1 = 1 / 2, a (n + 1) = an + 1 / (n square + n), find an

a(n+1)=an+1/[n(n+1)]
=>a(n+1)-an=1/[n(n+1)]=1/n-1/(n+1)
When n ≥ 1,
a2-a1=1-1/2
a3-a2=1/2-1/3
a4-a3=1/3-1/4
...
an-a(n-1)=1/(n-1)-1/n
a(n+1)-an=1/n-1/(n+1)
If the two sides of the above equation are added separately, there will be
(a2-a1)+(a3-a2)+(a4-a3)+...+[an-a(n-1)]+[a(n+1)-an]=(1-1/2)+(1/2-1/3)+(1/3-1/4)+...+[1/(n-1)-1/n]+[1/n-1/(n+1)]
=>a(n+1)-a1=1-1/(n+1)
=>a(n+1)=3/2-1/(n+1)
=>an=3/2-1/n

Given that the sequence {an} satisfies A1 = 1, an = a (n-1) Square-1 (n > 1), write its first five terms

1,0,-1,0,-1

1+3+7+15 +…… +(2n + 1), sum

[1+(2n+1)]n/2

Split term summation!
1+1/（1+2）+1/（1+2+3）+1/（1+2+3+4）+...+1/(1+2+...+n)=?
What matters is how to do it.

Original formula = 1 + 1 / 3 + 1 / 6 +... + 2 / [n (n + 1)]
=1+1/3+1/6+...+2[(1/n)-1/(n+1)]
=1+2[1/2-1/3+1/3-1/4+...+1/n-1/(n+1)]
=1+2[1/2-1/(n+1)]
=2-2/(n+1)
=2n/(n+1）
The key point is 1 / [n (n + 1)] = 1 / n-1 / (n + 1)

BN = 1 / (2n-1) (2n) sum

bn=1/[(2n-1)(2n)]=1/(2n-1)-1/(2n)
∴Sn=1/(1*2)+1/(3*4)+… +1/(2n-1)(2n)=1-1/2+1/3-1/4+1/5-1/6+… +1/(2n-1)-1/(2n)
When n →∞, Sn = ∑ BN = LN2

BN + 1 = BN + 2N-1 BN = - 1 finding the general term of BN

bn+1 - bn=2n-1
b2-b1=2*1-1
b3-b2=2*2-1
and so on
bn-bn-1=2(n-1)-1
accumulation
On the left, the total elimination is only bn-b1
The sum of all the items on the right is n-1

Sum n ^ 3 + (n + 1) ^ 3 + +(2n)^3
n^3+(n+1)^3+(n+2)^3…… +(2n)^3

There is a formula: 1 ^ 3 + 2 ^ 3 + 3 ^ 3 +. + n ^ 3 = [n (n + 1) / 2] ^ 2, so the original formula = [1 ^ 3 + 2 ^ 3 + 3 ^ 3 +. + (2n) ^ 3] - [1 ^ 3 + 2 ^ 3 + 3 ^ 3 +. + (n-1) ^ 3] = [(2n) (2n + 1) / 2] ^ 2 - [(n-1) n / 2] ^ 2 = (15N ^ 4 + 18N ^ 3 + 3N ^ 2) / 4 = 3 / 4 * n ^ 2 * (n + 1) * (5 * n + 1)

Sum split term cancellation (2 & # 178 / 1 * 3) + (4 & # 178 / 3 * 5) + (6 & # 178 / 5 * 7) +. + [(2n) & # 178; / (2n-1) * (2n + 1)]

（2²/1*3）+（4²/3*5）+（6²/5*7）+.+[﹙2n﹚²/（2n-1）*（2n+1）]
=n+1/1*3+1/3*5）+1/5*7+.+1/（2n-1）*（2n+1）
=n+[1-1/3+1/3-1/5+1/5-1/7+…… +1/(2n-1)-1/(2n+1)]÷2
=n+[1-1/(2n+1)]÷2
=n+2n/(2n+1)÷2
=n+n/(2n+1)

Sum 1 / (1 * 3) + 1 / (3 * 5) + 1 / (5 * 7) +... + 1 / ((2n-1) * (2n + 1))
Ask for detailed explanation!

1/(1*3)+1/(3*5)+1/(5*7)+...+1/((2n-1)*(2n+1))
=1/2[1-1/3+1/3-1/5+1/5-1/7+…… +1/(2n-1)-1/(2n+1)]
=1/2[1-1/(2n+1)]
=1/2*2n/(2n+1)
=n/(2n+1)