(1) prove that: 1 + 4 + 7 +………+ (3n – 2) = (3n – 1). ( n ≥ 1 )

The statement as given is not true. For n = 1, the left side is just the first term, 1, while the right side would be 3(1) - 1 = 2, and 1 ≠ 2.

I think what you meant to write was the equation

1 + 4 + 7 + ... + (3n - 2) = 1/2 n (3n - 1)

which can be easily proved by induction.

For n = 1, we have

1 = 1/2 (1) (3(1) - 1)   →   1 = 1

which is of course true.

Now for the inductive step. Assume the equation holds for n = k, so that

1 + 4 + 7 + ... + (3k - 2) = 1/2 k (3k - 1)

and use this to show it holds for n = k + 1.

We would have

1 + 4 + 7 + ... + (3k - 2) + (3(k + 1) - 2)

which reduces to

1/2 k (3k - 1) + (3k + 1)

3/2 k² + 5/2 k + 1

1/2 (3k² + 5k + 2)

1/2 (k + 1) (3k + 2)

which is precisely what the formula gives on the right side for n = k + 1. QED

uhhh idk u yolo?

hello from mrbilldoesmath!

answer:

25x^2 + 90x + 84

discussion:

mn(x) = m(n(x)) = m( 5x + 9) =

(5x + 9)^2 + 3 =>

25x^2 + 90x + 9^2 + 3 =

25x^2 + 90x + 84                 (as 9^2 = 9*9 = 81)

which is the second choice

regards,  

mrb

p.s.   i'll be on vacation from friday, dec 22 to jan 2, 2019. have a great new year!