Let a, b, c, and d, be positive integer constants with a < b. Without using the arithmetic sum formula, prove that (c(a + b) + 2d) (b – a + 1) > (ci + d) = %3D 2 i=a a-1 (ci + d) = ci - ci + d i=a i=1 i=1 i=a cb(b + 1) c(a– 1)a | + (b – a + 1)d c(b² + b– (a² – a)) 2d(b – a + 1) | 2 2 c(b? +b- a2 + a), 2d(b - a+ 1) + 2 2 c(b2 - a2 +b+ a) , 2d(b – a + 1) + c((b- a)(b + a) + (b + a)) , 2d(b - a + 1) + | 2 + a)(b- a + 1), 2d(b – a + 1) 2 (c(a + b) + 2d)(b – a + 1) 2
Let a, b, c, and d, be positive integer constants with a < b. Without using the arithmetic sum formula, prove that (c(a + b) + 2d) (b – a + 1) > (ci + d) = %3D 2 i=a a-1 (ci + d) = ci - ci + d i=a i=1 i=1 i=a cb(b + 1) c(a– 1)a | + (b – a + 1)d c(b² + b– (a² – a)) 2d(b – a + 1) | 2 2 c(b? +b- a2 + a), 2d(b - a+ 1) + 2 2 c(b2 - a2 +b+ a) , 2d(b – a + 1) + c((b- a)(b + a) + (b + a)) , 2d(b - a + 1) + | 2 + a)(b- a + 1), 2d(b – a + 1) 2 (c(a + b) + 2d)(b – a + 1) 2
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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![Let a, b, c, and d, be positive integer constants with a < b. Without using the arithmetic sum formula,
prove that
(c(a + b) + 2d) (b – a + 1)
>(ci + d)
%3D
2
i=a
a-1
> (ci + d) =
a+ο -Σά-Σ.
ci –
ci +
d
i=a
i=1
i=1
i=a
cb(b + 1) с(а — 1)а
+ (b – a + 1)d
c(b² + b – (a² – a)) 2d(b – a + 1)
2
2
c(b2 + b – a² +a) , 2d(b – a + 1)
-
2
2
c(b? – a² + b + a) , 2d(b – a + 1)
+
c((b – a)(b + a) + (b + a)) , 2d(b – a + 1)
+
2
2
с (b + a)(b — а+1)
2d(b — а + 1)
+
2
2
(c(a + b) + 2d)(b – a + 1)
2](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F7d4fd4d4-77c7-4e93-950b-25099f54e660%2Fdf59817d-3169-4b98-ab4e-9837671d19f7%2F42tmig_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Let a, b, c, and d, be positive integer constants with a < b. Without using the arithmetic sum formula,
prove that
(c(a + b) + 2d) (b – a + 1)
>(ci + d)
%3D
2
i=a
a-1
> (ci + d) =
a+ο -Σά-Σ.
ci –
ci +
d
i=a
i=1
i=1
i=a
cb(b + 1) с(а — 1)а
+ (b – a + 1)d
c(b² + b – (a² – a)) 2d(b – a + 1)
2
2
c(b2 + b – a² +a) , 2d(b – a + 1)
-
2
2
c(b? – a² + b + a) , 2d(b – a + 1)
+
c((b – a)(b + a) + (b + a)) , 2d(b – a + 1)
+
2
2
с (b + a)(b — а+1)
2d(b — а + 1)
+
2
2
(c(a + b) + 2d)(b – a + 1)
2
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