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Eduardwww [97]
3 years ago
7

Hey and I supposed to add or multiple when there is volume?​

Mathematics
2 answers:
MrMuchimi3 years ago
8 0

Answer:

mutiply

Step-by-step explanation:

gladu [14]3 years ago
3 0
Multiply hope this helps
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1. Find the sum of the following AP (i) 2, 4, 6, ... to n terms (ii) 7, 71, 72, 73, ... to 50 terms (iii) a, (a + b), (a + 2b),
Mrac [35]

Let S_n be the sum of the first n natural numbers.

S_n = 1 + 2 + 3 + \cdots + n

We can reverse the order of terms to write

S_n = n + (n-1) + (n-2) + \cdots + 1

so that doubling up, we get

2S_n = (1 + n) + (2 + (n-1)) + (3 + (n-2)) + \cdots + (n + 1)

2S_n = (n + 1) + (n + 1) + (n + 1) + \cdots + (n + 1)

and since there are n terms in S_n, it follows that

2S_n = n(n+1) \implies S_n = \dfrac{n(n+1)}2

Now, for an arithmetic sequence starting with a_0 and having common difference d between consecutive terms, the sum of the first n terms of this sequence is

a_0 + (a_0 + d) + (a_0 + 2d) + \cdots + (a_0 + (n-1)d) \\\\ ~~~~~~~~ = (a_0 - d + d) + (a_0 - d + 2d) + (a_0 - d + 3d) + \cdots + (a_0 - d + nd) \\\\ ~~~~~~~~ = (a_0 - d) n + d S_n \\\\ ~~~~~~~~ = \dfrac d2 n^2 + \left(a_0 - \dfrac d2\right) n

(i) This sum is simply

2 + 4 + 6 + \cdots + 2n \\\\ ~~~~~~~~ = 2 (1 + 2 + 3 + \cdots + n) \\\\ ~~~~~~~~ = 2 S_n \\\\ ~~~~~~~~ = \boxed{n(n+1)}

(ii) The first term in this sequence should be 70 if it's arithmetic. The common difference is then 1, the 50th term is 70 + (50 - 1)•1 = 119, so

70 + 71 + 72 + \cdots + 119 \\\\ ~~~~~~~~ = \dfrac12\cdot50^2 + \left(70 - \dfrac12\right)\cdot50 \\\\ ~~~~~~~~ = \boxed{4725}

(iii) The first term is a and the common difference is b, so

a + (a + b) + (a + 2b) + \cdots + (a + b(n-1)) \\\\ ~~~~~~~~ = \boxed{\dfrac b2 n^2 + \left(a - \dfrac b2\right)n}

(iv) The first term is x+y and the common difference is -2y. Then the 20th term in the sequence is x+y - 19\cdot2y= x-37y, so

(x + y) + (x - y) + (x - 3y) + \cdots + (x - 37y) \\\\ ~~~~~~~~ = -\dfrac{2y}2\cdot20^2 + \left(x + y + \dfrac{2y}2\right)\cdot20 \\\\ ~~~~~~~~ = \boxed{20x - 360y}

(v) Note that the first term is

(a - b)^2 = a^2 + b^2 - 2ab

so that the common difference is 2ab. Then

(a - b)^2 + (a^2 + b^2) + (a + b)^2 + \cdots + (a^2 + b^2 - 2ab + 2ab(n-1)) \\\\ ~~~~~~~~ = \dfrac{2ab}2 n^2 + \left(a^2 + b^2 - 2ab - \dfrac{2ab}2\right) n \\\\ ~~~~~~~~ = \boxed{abn^2 + (a^2 - 3ab + b^2)n}

7 0
2 years ago
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