Solve by Arithmetic Progression 1.Find the 11th term of 2,4,6… 2. Find the sum 1+3+5+…(2n -1)?
回答 (3)
2020-08-13 12:13 am
✔ 最佳答案
1.First term, a = 2
Common difference, d = 4 - 2 = 2
11th term
= a + (n - 1) d
= 2 + (11 - 1) × 2
= 22
====
2.
First term, a = 1
Common difference = 3 - 1 = 2
No. of terms = n
Sum of n terms
= n [2a + (n - 1) d] / 2
= n [2(1) + (n - 1) × 2] / 2
= n [2 + 2n - 2)] / 2
= n (2n) / 2
= n²
2020-08-12 11:57 pm
1) 2, 4, 6,..., 2n
so, 11th term is 2(11) = 22
2) ∑(2n - 1) => ∑2n - n
i.e. 2∑n - n
Now, ∑n => n(n + 1)/2
so, 2∑n => n(n + 1)
Hence, n(n + 1) - n => n²
i.e. the sum of the first n odd numbers is the nth square number.
:)>
so, 11th term is 2(11) = 22
2) ∑(2n - 1) => ∑2n - n
i.e. 2∑n - n
Now, ∑n => n(n + 1)/2
so, 2∑n => n(n + 1)
Hence, n(n + 1) - n => n²
i.e. the sum of the first n odd numbers is the nth square number.
:)>
2020-08-13 12:18 am
1.
The 11th term of 2, 4, 6, 8, … is 22.
2.
a_n = 2 n - 1
1 + 3 + 5 +… + 39 = 400
The 11th term of 2, 4, 6, 8, … is 22.
2.
a_n = 2 n - 1
1 + 3 + 5 +… + 39 = 400
收錄日期: 2021-04-24 07:58:41
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20200812154250AAAwPLQ