Question

Find the first six terms of the sequence.

a1 = -6, an = 4 • an-1

0, 4, -24, -20, -16, -12
-24, -96, -384, -1536, -6144, -24,576
-6, -24, -20, -16, -12, -8
-6, -24, -96, -384, -1536, -6144

Answer 1

i know its either c or d ..... def. not a or b because it clearly tells you a1 = -6

Answer 2

@akonkel @idku

Answer 3

d

Answer 4

an=4*an-1 right?

that means -24=-6*4
-96=-24*4
etc etc

Answer 5

yeah okay i see that now.. thank you (: wanna help with another?

Answer 6

sure

Answer 7

Use mathematical induction to prove the statement is true for all positive integers n.

The integer n3 + 2n is divisible by 3 for every positive integer n.

Answer 8

is that n^3 or 3n?

Answer 9

n^3

Answer 10

( n + 1 )^3 + 2( n + 1 ) = ( n^3 + 3n^2 + 3n + 1 ) + ( 2n + 2 )
= ( n^3 + 2n ) + ( 3n^2 + 3n + 3 )
= ( n^3 + 2n ) + 3( n^2 + n + 1 )
Therefore ( n^3 + 2n ) is divisible by 3
right?

Answer 11

im actually not familiar with this sorry.

Answer 12

what about this one ?

Write the sum using summation notation, assuming the suggested pattern continues.

64 + 81 + 100 + 121 + ... + n2 + ...

summation of n squared from n equals eight to infinity
summation of n minus one squared from n equals eight to infinity
summation of n squared from n equals nine to infinity
summation of n plus one squared from n equals eight to infinity

Answer 13

@TylerD

Answer 14

looks like the first one

Answer 15

n^2

starting at n=8

so it starts at 8^2 which = 64

then the next term would be 9^2 = 81

next term 10^2 = 100 etc etc etc