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7.04

Use mathematical induction to prove the statement is true for all positive integers n, or show why it is false.
(4 points each.)

1. 4 ⋅ 6 + 5 ⋅ 7 + 6 ⋅ 8 + ... + 4n( 4n + 2) = quantity four times quantity four n plus one times quantity eight n plus seven divided all divided by six


2. 12 + 42 + 72 + ... + (3n - 2)2 = quantity n times quantity six n squared minus three n minus one all divided by two


For the given statement Pn, write the statements P1, Pk, and Pk+1.
(2 points)

2 + 4 + 6 + . . . + 2n = n(n+1)

Respuesta :

kerriamolden
4 ⋅ 6 + 5 ⋅ 7 + 6 ⋅ 8 + ... + 4n( 4n + 2) = 4(4n+1)(8n+7)/6 is false because it isn't even true when n=1. 4 ⋅ 6 = 4(4*1+1)(8*1+7)/6 24 = 4(5)(15)/6 24 = 50 Also 4n(4n+2) is not even the correct formula for the nth term. The correct nth term formula of 4 ⋅ 6 + 5 ⋅ 7 + 6 ⋅ 8 + ... is (n+3)(n+5). So it's wrong all the way around.
lublana

Answer:

1. False.

2.False.

3.[tex]P_{1} =[/tex]2

[tex]P_{k}[/tex]=2+4+6+.....+2n= n(n+1)

[tex]P_{k+1}[/tex]=2+4+6+....+2(k+1)=(k+1)(k+2)

Step-by-step explanation:

1.Given

4.6+5.7+6.8+.....+4n(4n+2)=[tex]\frac{4(4n+1)(8n+7)}{6}[/tex]

First , we prove for n=1

[tex]P_{1}[/tex]= 4.6=24

Left hand side :

[tex]P_{1}[/tex]=[tex]4\times1(4\times1+2)[/tex]

[tex]P_{1}[/tex]=24

Right hand side:

[tex]P_{n}[/tex]=[tex]\frac{4(4n+1)(8n+7)}{6}[/tex]

Put n=1

[tex]P_{1}[/tex]=[tex]\frac{4(4\times1 +1)(8\times1 +7)}{6}[/tex]

[tex]P_{1}[/tex]= [tex]\frac{4(4\times1 +1)(8\times1+7)}{6}[/tex]

[tex]P_{1}[/tex]= [tex]\frac{4\times5\times15}{6}[/tex]

[tex]P_{1}[/tex]=50

Hence, left hand side≠ right hand side

Therefore, the given statement is false.

2. [tex]P_{n}[/tex]=12+42+72+.......(3n-2)2=[tex]\frac{n(6n^{2-3n-1)} }{2}[/tex]

similarly , we prove that in the same manner

First , we prove for n=1 the given statement is true or false.

Take n=1

[tex]P_{1}[/tex]=12

Left hand side :

[tex]P_{n}[/tex]=(3n-2)2

Put n=1 then

[tex]P_{1}[/tex]=[tex](3\times1-2)2[/tex]

[tex]P_{1}[/tex]=2

Right hand side:

[tex]P_{n} =[tex]\frac{n(6n^{2-3n-1)} }{2}[/tex]

Put n=1

Then we get

[tex]P_{1}[/tex]=[tex]\frac{1(6\times1-3\times1-1)}{2}[/tex]

[tex]P_{1}[/tex]=[tex]\frac{4}{2}[/tex]

[tex]P_{1}[/tex]=2

Hence , left hand side≠right hand side ≠[tex]P_1[/tex]

Therefore, the given statement is false.

3. Given that

[tex]P_{n}[/tex]=2+4+6+....+2n=n(n+1)

Put n= 1

[tex]P_{1}[/tex]=2

Put n=k then we get the statement  Pk

[tex]P_{k}[/tex]=2+4+6+....+2k=k(k+1)

Now , Put n= k+1 and then we get the statment Pk+1

[tex]P_{k+1}[/tex]=2+4+6+....+2(k+1)=(k+1)(k+2).

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