"The sum of three consecutive natural numbers is by N N greater than thrice the arithmetic average of the smallest and the largest of the three numbers."

Answer 1

X + X+1 + X+2 = 3 ( x + x+2)/2 + N

3x + 3 = 3(2x + 2) /2 + N <--- combines like terms

3(x+1) = 3* 2(x+1)/2 + N <--- factors out GCF of 3 and 2

3(x+1) = 3*(x+1) + N <--- you can factor out the 2 and divide it out;

division higher priority than addition PEMDAS

0 = N <--- subtracts 3(x+1) from both sides

So there is no difference between the sum of the integers and the average of the extremes.

the equation becomes

X + X+1 + X+2 = 3 ( x + x+2)/2

3x + 3 = 3(2x +2)/2

3(x+1) = 3(2)(x+1)/2

3(x+1) = 3(x+1)

So the identity holds for all integers.

This can be proven by induction...

x=1:

1 + 2 + 3 = 6 = 12/2 = 3(4)/2 = 3( 1+ 3)/2 <--- the pattern holds for x = 1

x=2:

2+3+4 = 9 = 3*3 = 3*6/2 = 3*(2+4)/2 <--- the pattern holds for x=2

Suppose for some positive integer k:

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

Then

3 ( k+1 + k+3) /2 = 3 ( 2k + 4)/2 = 3(2)(k+2)/2 = 3(k+2) = 3k + 6

= k+k+k+ 1+2+3

= (k+1)+(k+2)+(k+3)

Answer 2

What does this mean: "is by N N greater "

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