Schmidtjoseph174
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Use mathematical induction to prove the statement is true for all positive integers n, or show why it is false.
(4 points each.)

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





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 :

1. The given series is

 4 ⋅ 6 + 5 ⋅ 7 + 6 ⋅ 8 + ... + 4n( 4n + 2) =  

For n=1

L.H.S=4.6=24

R.H.S=[4×5×15]÷6

      =300÷6

      =50

So, for n=1,

L.H.S≠ R.H.S

Since the given expression is true for n=1 ,

So , the given series is untrue.

we should replace R.H.S by=4(n+1)(n+2)(4n-3)²

2.

12+42+72+.......+(3 n -2)2=

For n=1,

L.H.S=12

R.H.S=1×(6-3-1)/2

       =2/2

      =1

As L.H.S≠ R.H.S

We should Replace R.H.S by [(3 n-1)(3 n-2)]2

3.The given sequence is

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

L.H.S

P(1)=2

R.H.S

1×(1+1)

=1×2

=2

( b)  L.H.S

P(n)=2+4+6+.....2 k

This is an A.P having n terms.

tex]S_{n}=

              = n(n+ 1)

R.H.S=n(n+1)

So, P(k)=k(k+1)

(c) P(k+1)=2+4+6+.......+2(k+1)

This is an A.P having (k+1) terms.

              =(k+1)(k+2)

So, P(k+1)= (k+1)(k+2)

(Btw this is not my answer)

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