Given any real numbers a and d, consider the expression (a + md) + (a + (m + 1)d) + (a + (m + 2)d) + + (a + (m + n)d). Show that if m and n are any integers with n ≥ 0, then 1. (a + md) + (a + (m + 1)d) + (a + (m + 2)d) + 2. (a + md) + (a + (m + 1)d) + (a + (m + 2)d) + Now 3. (a + md) + (a + (m + 1)d) + (a + (m + 2)d) ++ (a + (m + n)d) = Proof: Suppose a and d are any real numbers and m and n are any integers with n 20. Then (a + md) + (a + (m + 1)d) + (a + (m + 2)d) + ... + (a + (m + n)d) above expression contains (a + md) + (a + md + d) + (a + md + 2d)+...+(a + md + nd) by multiplying out Observe that each term in the above expression contains (a + md), and the first (a + md) can be written as (a + md + 0d). Since the last term is (a + md + nd), there are as many terms as there are numbers from 0 through n, namely Thus, by the transitivity of equality (a + md) + (a + md + d) + (a + md + 2d) + ... + (a + m + nd) = (a + md) + (a + md) + + (a + (m + n)d) = (n + 1)(a + md) + copies of d. Hence, where there are (n + 1) terms of (a + md). Then (a + md) + (a + md) ++ (a + md) + (d+ 2d + 3d++ nd) = (n + 1)(a + md) + d(1 + 2 + 3+ ... + n) = (n + 1)(a + md) + d + d(n(n + 1)) + (a + (m + n)d) = (a + md + (n + 1)(a + md) + d ) + d( ^(^ + ¹) ) = ( Finally, (n+1) ✓, we can conclude that Formula (1) is true. (1) (a + md) + (a + (m + 1)d) + (a + (m + 2)d) + + (a + (m + n)d) = (n + 1)(a + md) + d(n(n + ¹)) 2 = (n + 1a + md + d- n+1)/(² + d²/-) = (a + md + 2 d) (n + 1 + 2/d)(n+1) = [₁ + (m + ²/7) α)](n + 1) + (a + md) + (d+ 2d 3d + + nd), Hence, Formula (2) is true by factoring out a common factor + 1){( a + md + d² ) = [a + (m + ½ ) o](n + ² + d(n(n + ¹)) where there are (n + 1) terms of (a + md) by factoring out a common factor ✓ ✓✔ by Theorem 5.2.1 by multiplying out (n + 1) by multiplying out P ✓ V and thus Formula (3) is true by Theorem 5.2.1 In addition, the
Given any real numbers a and d, consider the expression (a + md) + (a + (m + 1)d) + (a + (m + 2)d) + + (a + (m + n)d). Show that if m and n are any integers with n ≥ 0, then 1. (a + md) + (a + (m + 1)d) + (a + (m + 2)d) + 2. (a + md) + (a + (m + 1)d) + (a + (m + 2)d) + Now 3. (a + md) + (a + (m + 1)d) + (a + (m + 2)d) ++ (a + (m + n)d) = Proof: Suppose a and d are any real numbers and m and n are any integers with n 20. Then (a + md) + (a + (m + 1)d) + (a + (m + 2)d) + ... + (a + (m + n)d) above expression contains (a + md) + (a + md + d) + (a + md + 2d)+...+(a + md + nd) by multiplying out Observe that each term in the above expression contains (a + md), and the first (a + md) can be written as (a + md + 0d). Since the last term is (a + md + nd), there are as many terms as there are numbers from 0 through n, namely Thus, by the transitivity of equality (a + md) + (a + md + d) + (a + md + 2d) + ... + (a + m + nd) = (a + md) + (a + md) + + (a + (m + n)d) = (n + 1)(a + md) + copies of d. Hence, where there are (n + 1) terms of (a + md). Then (a + md) + (a + md) ++ (a + md) + (d+ 2d + 3d++ nd) = (n + 1)(a + md) + d(1 + 2 + 3+ ... + n) = (n + 1)(a + md) + d + d(n(n + 1)) + (a + (m + n)d) = (a + md + (n + 1)(a + md) + d ) + d( ^(^ + ¹) ) = ( Finally, (n+1) ✓, we can conclude that Formula (1) is true. (1) (a + md) + (a + (m + 1)d) + (a + (m + 2)d) + + (a + (m + n)d) = (n + 1)(a + md) + d(n(n + ¹)) 2 = (n + 1a + md + d- n+1)/(² + d²/-) = (a + md + 2 d) (n + 1 + 2/d)(n+1) = [₁ + (m + ²/7) α)](n + 1) + (a + md) + (d+ 2d 3d + + nd), Hence, Formula (2) is true by factoring out a common factor + 1){( a + md + d² ) = [a + (m + ½ ) o](n + ² + d(n(n + ¹)) where there are (n + 1) terms of (a + md) by factoring out a common factor ✓ ✓✔ by Theorem 5.2.1 by multiplying out (n + 1) by multiplying out P ✓ V and thus Formula (3) is true by Theorem 5.2.1 In addition, the
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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![Given any real numbers a and d, consider the expression
(a + md) + (a + (m + 1)d) + (a + (m + 2)d) + ... + (a + (m + n)d).
Show that if m and n are any integers with n ≥ 0, then
1. (a + md) + (a + (m +
2. (a + md) + (a + (m +
3. (a + md) + (a + (m + 1)d) + (a + (m + 2)d) + ... + (a + (m + n)d) = a + - (m + ²/2) α] (n +
+ 1).
above expression contains
Proof: Suppose a and d are any real numbers and m and n are any integers with n ≥ 0. Then
(a + md) + (a + (m + 1)d) + (a + (m + 2)d) + ... + (a + (m + n)d)
= (a + md) + (a + md + d) + (a + md + 2d) + ... + (a + md + nd) by multiplying out
1)d) + (a + (m + 2)d) + ... + (a + (m + n)d) = (n + 1)(a + md) + d
Observe that each term in the above expression contains (a + md), and the first (a + md) can be written as (a + md + 0d). Since the last term is (a + md + nd), there are as many terms as there are numbers from 0 through n, namely
(a + md) + (a + md + d) + (a + md + 2d) + ... + (a + m + nd)
= (a + md) + (a + md) + ... + (a + md) + (d + 2d + 3d + ... + nd),
where there are (n + 1) terms of (a + md).
Now
1)d) + (a + (m + 2)d) + ... + (a + (m + n)d) = (a + md +
Then
(a + md) + (a + md) + ... + (a + md) + (d + 2d + 3d + ... + nd)
= (n + 1)(a + md) + d(1 + 2 + 3 + ... + n)
Thus, by the transitivity of equality
(n + 1)(a + md) +
copies of d. Hence,
= (n + 1)(a + md) + d
✓, we can conclude that Formula (1) is true.
(1) (a + md) + (a + (m + 1)d) + (a + (m + 2)d) + ・・・ + (a + (m + n)d)
+ a(n(n+1))
2
= (n + 1)(a + md) + d(
n(n+1)
1 + d( ~²n + ¹)) = (n + 1)(a + ma + ²)
md d
Finally, (n + 1) a + md +
n(n + 1)
2
Hence, Formula (2) is true by factoring out
=(2+ một nơi + 1)
(² 2/d) (n.
(n
common factor .
+ d²2²) = [2 + ( m + ²27) d]‹n +
+ 1/2d) (n +
(n + 1)
where there are (n + 1) terms of (a + md)
by factoring out a common factor ✓ ✓
by Theorem 5.2.1
by multiplying out
( n(n + ¹))
2
(n + 1) by multiplying out
V
v and thus Formula (3) is true by Theorem 5.2.1
In addition, the](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F9824ebe1-81ec-4540-832f-9b2c74c8a4d5%2F41533a90-f35b-4982-a1ab-07e05e8e48d7%2Fuaxatzk_processed.png&w=3840&q=75)
Transcribed Image Text:Given any real numbers a and d, consider the expression
(a + md) + (a + (m + 1)d) + (a + (m + 2)d) + ... + (a + (m + n)d).
Show that if m and n are any integers with n ≥ 0, then
1. (a + md) + (a + (m +
2. (a + md) + (a + (m +
3. (a + md) + (a + (m + 1)d) + (a + (m + 2)d) + ... + (a + (m + n)d) = a + - (m + ²/2) α] (n +
+ 1).
above expression contains
Proof: Suppose a and d are any real numbers and m and n are any integers with n ≥ 0. Then
(a + md) + (a + (m + 1)d) + (a + (m + 2)d) + ... + (a + (m + n)d)
= (a + md) + (a + md + d) + (a + md + 2d) + ... + (a + md + nd) by multiplying out
1)d) + (a + (m + 2)d) + ... + (a + (m + n)d) = (n + 1)(a + md) + d
Observe that each term in the above expression contains (a + md), and the first (a + md) can be written as (a + md + 0d). Since the last term is (a + md + nd), there are as many terms as there are numbers from 0 through n, namely
(a + md) + (a + md + d) + (a + md + 2d) + ... + (a + m + nd)
= (a + md) + (a + md) + ... + (a + md) + (d + 2d + 3d + ... + nd),
where there are (n + 1) terms of (a + md).
Now
1)d) + (a + (m + 2)d) + ... + (a + (m + n)d) = (a + md +
Then
(a + md) + (a + md) + ... + (a + md) + (d + 2d + 3d + ... + nd)
= (n + 1)(a + md) + d(1 + 2 + 3 + ... + n)
Thus, by the transitivity of equality
(n + 1)(a + md) +
copies of d. Hence,
= (n + 1)(a + md) + d
✓, we can conclude that Formula (1) is true.
(1) (a + md) + (a + (m + 1)d) + (a + (m + 2)d) + ・・・ + (a + (m + n)d)
+ a(n(n+1))
2
= (n + 1)(a + md) + d(
n(n+1)
1 + d( ~²n + ¹)) = (n + 1)(a + ma + ²)
md d
Finally, (n + 1) a + md +
n(n + 1)
2
Hence, Formula (2) is true by factoring out
=(2+ một nơi + 1)
(² 2/d) (n.
(n
common factor .
+ d²2²) = [2 + ( m + ²27) d]‹n +
+ 1/2d) (n +
(n + 1)
where there are (n + 1) terms of (a + md)
by factoring out a common factor ✓ ✓
by Theorem 5.2.1
by multiplying out
( n(n + ¹))
2
(n + 1) by multiplying out
V
v and thus Formula (3) is true by Theorem 5.2.1
In addition, the
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