Question

PLEASE HELP AND SHOW ALL WORK

Answer 1

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.

Answer 2

Answer:

1. False.

2.False.

3.$P_{1} =$2

$P_{k}$=2+4+6+.....+2n= n(n+1)

$P_{k+1}$=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)=$\frac{4(4n+1)(8n+7)}{6}$

First , we prove for n=1

$P_{1}$= 4.6=24

Left hand side :

$P_{1}$=$4\times1(4\times1+2)$

$P_{1}$=24

Right hand side:

$P_{n}$=$\frac{4(4n+1)(8n+7)}{6}$

Put n=1

$P_{1}$=$\frac{4(4\times1 +1)(8\times1 +7)}{6}$

$P_{1}$= $\frac{4(4\times1 +1)(8\times1+7)}{6}$

$P_{1}$= $\frac{4\times5\times15}{6}$

$P_{1}$=50

Hence, left hand side≠ right hand side

Therefore, the given statement is false.

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

similarly , we prove that in the same manner

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

Take n=1

$P_{1}$=12

Left hand side :

$P_{n}$=(3n-2)2

Put n=1 then

$P_{1}$=$(3\times1-2)2$

$P_{1}$=2

Right hand side:

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

Put n=1

Then we get

$P_{1}$=$\frac{1(6\times1-3\times1-1)}{2}$

$P_{1}$=$\frac{4}{2}$

$P_{1}$=2

Hence , left hand side≠right hand side ≠$P_1$

Therefore, the given statement is false.

3. Given that

$P_{n}$=2+4+6+....+2n=n(n+1)

Put n= 1

$P_{1}$=2

Put n=k then we get the statement  Pk

$P_{k}$=2+4+6+....+2k=k(k+1)

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

$P_{k+1}$=2+4+6+....+2(k+1)=(k+1)(k+2).