Let p(n): 23 27 = 2 - (n+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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In question 10 part be , in the solution part what does p(n) mean please
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п(2n - 1)(2n + 1)
for all n > 1.
(d) 1² + 3² + 5² + … … - + (2n – 1)² =
3
ding
(e) 13 + 2³ + 3³ + ..+n³ =
for all n > 1.
2 fr + 1, show that for any positive integer n,
a(n+1 – 1)
a + ar + ar“ +… . + ar" =
r - 1
3. Use the Second Principle of Finite Induction to establish that for all n > 1,
a" – 1 = (a – 1)(a"-1 +a"-2 + a"-3+ ..+ a + 1)
[Hint: a"+1 – 1 = (a + 1)(a" – 1) – a(a"-1 – 1).]
4 Prove that the cube of any integer can be written as the difference of two squares. [Hint:
Notice that
the
ive
n² = (1³ + 2³ + . . + n³) – (1³ + 2³ + • … + (n – 1)').]
5. (a) Find the values of n < 7 for which n! +1 is a perfect square (it is unknown whether
n! +1 is a square for any n > 7).
(b) True or false? For positive integers m and n, (mn)! = m!n!and (m + n)! = m! + n!.
6. Prove that n! > n² for every integer n > 4, whereas n! > n³ for every integer n > 6.
T.Use mathematical induction to derive the following formula for all n > 1:
Dr
le
1(1!) + 2(2!) + 3(3!) + · . · + n(n!) = (n+1)! – 1
8. (a) Verify that for all n > 1,
(2n)!
2.6 10 14 ..... (4n – 2) =
n!
(b) Use part (a) to obtain the inequality 2" (n!)² < (2n)! for all n > 1.
9. Establish the Bernoulli inequality: If 1 + a > 0, then
(1+ a)" > 1+ na
for all n > 1.
10.) For all n > 1, prove the following by mathematical induction:
1
1
<2
(a)
12
22
32
n2
n
1
3
n+2
...+
= 2
22 23
2n
2n
11. Show that the expression (2n)!/2"n! is an integer for all n > 0.
12. Consider the function defined by
3n +1
for n odd
T(n) =
for n even
conjecture is the claim that starting from any integer n > 1, th
T(n), T(T (n)), T(T(T (n)), .. , eventually reaches the integer 1 a.
runs through the values 1 and 2. This has been verified for all n < 1016. CoOl](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F66c7b497-dac7-4855-b923-2e60bbc73063%2Ffe295bbf-21ef-4b1f-af46-2376cf09ff89%2F5thgw2t_processed.png&w=3840&q=75)
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п(2n - 1)(2n + 1)
for all n > 1.
(d) 1² + 3² + 5² + … … - + (2n – 1)² =
3
ding
(e) 13 + 2³ + 3³ + ..+n³ =
for all n > 1.
2 fr + 1, show that for any positive integer n,
a(n+1 – 1)
a + ar + ar“ +… . + ar" =
r - 1
3. Use the Second Principle of Finite Induction to establish that for all n > 1,
a" – 1 = (a – 1)(a"-1 +a"-2 + a"-3+ ..+ a + 1)
[Hint: a"+1 – 1 = (a + 1)(a" – 1) – a(a"-1 – 1).]
4 Prove that the cube of any integer can be written as the difference of two squares. [Hint:
Notice that
the
ive
n² = (1³ + 2³ + . . + n³) – (1³ + 2³ + • … + (n – 1)').]
5. (a) Find the values of n < 7 for which n! +1 is a perfect square (it is unknown whether
n! +1 is a square for any n > 7).
(b) True or false? For positive integers m and n, (mn)! = m!n!and (m + n)! = m! + n!.
6. Prove that n! > n² for every integer n > 4, whereas n! > n³ for every integer n > 6.
T.Use mathematical induction to derive the following formula for all n > 1:
Dr
le
1(1!) + 2(2!) + 3(3!) + · . · + n(n!) = (n+1)! – 1
8. (a) Verify that for all n > 1,
(2n)!
2.6 10 14 ..... (4n – 2) =
n!
(b) Use part (a) to obtain the inequality 2" (n!)² < (2n)! for all n > 1.
9. Establish the Bernoulli inequality: If 1 + a > 0, then
(1+ a)" > 1+ na
for all n > 1.
10.) For all n > 1, prove the following by mathematical induction:
1
1
<2
(a)
12
22
32
n2
n
1
3
n+2
...+
= 2
22 23
2n
2n
11. Show that the expression (2n)!/2"n! is an integer for all n > 0.
12. Consider the function defined by
3n +1
for n odd
T(n) =
for n even
conjecture is the claim that starting from any integer n > 1, th
T(n), T(T (n)), T(T(T (n)), .. , eventually reaches the integer 1 a.
runs through the values 1 and 2. This has been verified for all n < 1016. CoOl
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Solution-
Let p(n): +
3
+
27
= 2 - (n+2)
Step 2
For n: 1,
L.H.S = 1
R.H.S- 2- ( 1 +2) _ 2-
ニ
.. L'H.S = R.H.S
p(m) is taue for n= 1.
ASsume that
P(m) is toue for n=K,
PIK):
늘 +
2.
3
2-(K+2)
ス
22
23 +
2K
四
Step 3
レ
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