Sum of sequence (1 / 1 * 3) + (1 / 3 * 5) + (1 / 5 * 7) +... + (1 / (2n-1) (2n + 1)) ( 1/1*3)+(1/3*5)+(1/5*7)+...+(1/(2n-1)(2n+1)) The final answer is the original

Sum of sequence (1 / 1 * 3) + (1 / 3 * 5) + (1 / 5 * 7) +... + (1 / (2n-1) (2n + 1)) ( 1/1*3)+(1/3*5)+(1/5*7)+...+(1/(2n-1)(2n+1)) The final answer is the original

an =1/(2n-1)(2n+1) =1/2[1/(2n-1)-1/(2n+1)] Sn =1/2[1-1/3]+1/2[1/3-1/5]+1/2[1/5-1/7]+...+1/2[1/(2n-3)-1/(2n-1)]+1/2[1/(2n-1)-1/(2n+1)] =1/2[1-1/3+1/3-1/5+1/5-1/7+...+1/(2n-3)-1/(2n-1)+1/(2n-1)-1/(2n+1)...

Sequence 1,3,6,10,15,21
What is the nth number expressed by N

It can be seen from the items given
A (n + 1) - A (n) = n + 1, then
a(n)-a(n-1)=n
……
a(2)-a(1)=2
Add the above to get
a(n+1)-a(1)=(n+1)+n+…… +2
So a (n + 1) = (n + 1) + N + +2+1
Then a (n) = n + +2+1=n*(n+1)/2

Sequence 1.3.6.10.15.21., item 100 is ()

1+2+3+...+99+100=5050

For the arithmetic sequence with odd items, the sum of odd items is 44, and the sum of even items is 33

Let the middle term be x and the number of terms be n,
x﹙n+1﹚／2＝44
x﹙n－1﹚／2=33
x=11,n=7

If the arithmetic sequence has 2n + 1 terms (n ∈ n *), and the sum of odd and even terms is 44 and 33, then the number of terms is ()
A. 5B. 7C. 9D. 11

S = a1 + a3 + +A2N + 1 = (n + 1) (a1 + A2N + 1) 2 = (n + 1) an + 1, s-even = A2 + A4 + A6 + +A 2n = n (a 2 + a 2n) 2 = Nan + 1, N + 1n = 4433, the solution is n = 3, the number of terms 2n + 1 = 7

If the number of terms of the arithmetic sequence an is odd, the sum of the odd terms is 44, and the sum of the even terms is 33, then the middle term and the number of terms are

Suppose that there are 2n + 1 items in the arithmetic sequence, and the tolerance is D, then the even item is the arithmetic sequence of tolerance 2D, and there are n items, the odd item is the arithmetic sequence of tolerance 2D, and there are n + 1 items, and the middle item is a (n + 1)
N = (nadn) = (1) + n (1) + 1
S odd = (n + 1) a1 + 2DN (n + 1) / 2 = (n + 1) a1 + DN & # 178; + DN = 44 (2)
(2)-(1)
a1+nd=11
a(n+1)=11
S=a1+a2+...+a(2n+1)
=[a1+a(2n+1)]+[a2+a(2n)]+...+[an+a(n+2)]+a(n+1)
=2a(n+1)+2a(n+1)+...+2a(n+1)+a(n+1)
=(2n+1)a(n+1)
=11(2n+1)=44+33=77
2n+1=7
n=3
The middle item is 11 and the number of items is 3

In {an}, the sum of odd items is 80 and the sum of even items is 75
Find the middle item and number of items in the number column!

For the arithmetic sequence with odd terms, {an} has the following properties:
S odd-s even = in a, s odd-s even = (number of terms) * in a
So: the middle term is: 80-75 = 5
The number of items is: (80 + 75) / 5 = 31

In {an}, the sum of odd items is 80, and the sum of even items is 75

If the number of items is odd, we may as well set 2K + 1 items, then there are K + 1 odd items and K even items
Sum of odd terms = (a1 + a (2k + 1)) / 2 * (K + 1) = 80
Sum of even terms = (A2 + a (2k)) / 2 * k = 75
Where a1 + a (2k + 1) = A2 + a (2k)
So the solution is (a1 + a (2k + 1)) / 2 = (A2 + a (2k)) / 2 = 5, k = 15
So there are 31

It is known that an arithmetic sequence has 2n + 1 terms, and the sum of odd terms is 96, and the sum of even terms is 80
How to think about it

There are 2n + 1 items in the sequence, the first item is A1, and the tolerance is D. the odd item has n + 1 item, the even item has n item, and the middle item is the N + 1 item
Sum of odd terms S1 = (n + 1) [A1 + A1 + 2nd] / 2 = (n + 1) (a1 + nd)
Sum of even terms S2 = n [A1 + D + (a1 + D) + 2 (n-1) D] / 2 = n (a1 + nd)
S1-S2=a1+nd=a(n+1)=96-80=16
So the middle term is 16
Because a (1) + a (2n + 1) = 2 * a (n + 1), a (3) + a (2n-1) = 2 * a (n + 1)
a(1)+a(3)+...+a(2n+1)=(n+1)*a(n+1)
Similarly, a (2) + a (4) +... + a (2n) = n * a (n + 1)
So: (n + 1) / N = 96 / 80 = 6 / 5
n=5

The sum of odd terms of arithmetic sequence is 99, the sum of even terms is 90, the first term is a total term, the number of odd terms is the general term formula
The answer is 21

If the first item is 1 and the number of items is odd, then the odd item is more than the even item; if the even item has n items, then the odd item has n + 1 items; if the tolerance is D;
Then the tolerance of even and odd items is 2D;
There are: (1 + 2 × n × d) × (n + 1) / 2 = 99;
[1+d+（2×n-1）×d]×n/2=90;
The solution is n = 10;
d=17/20；