answer is C
He made an error in Step 3. He should have subtracted 2 from both sides.
Proof by induction
Base case:
n=1: 1*2*3=6 is obviously divisible by six.
Assumption: For every n>1 n(n+1)(n+2) is divisible by 6.
For n+1:
(n+1)(n+2)(n+3)=
(n(n+1)(n+2)+3(n+1)(n+2))
We have assumed that n(n+1)(n+2) is divisble by 6.
We now only need to prove that 3(n+1)(n+2) is divisible by 6.
If 3(n+1)(n+2) is divisible by 6, then (n+1)(n+2) must be divisible by 2.
The "cool" part about this proof.
Since n is a natural number greater than 1 we can say the following:
If n is an odd number, then n+1 is even, then n+1 is divisible by 2 thus (n+1)(n+2) is divisible by 2,so we have proved what we wanted.
If n is an even number" then n+2 is even, then n+1 is divisible by 2 thus (n+1)(n+2) is divisible by 2,so we have proved what we wanted.
Therefore by using the method of mathematical induction we proved that for every natural number n, n(n+1)(n+2) is divisible by 6. QED.
I believe the answer to your question is D
Answer: d=5
Step-by-step explanation:
Arithmetic progression
An=a1+d(n-1)
An= nth term
A1= first term
D= common difference
N= nth position
We are given the 4th and 7th terms
For 4th term,n=4
A4=a1+d(4-1)
A4= 16
16= a1+d(4-1)
16= a1+3d........equation 1
For the 7th term,n=7
A7=a1+d(7-1)
A7= 31
31=a1+6d......equation 2
Bring them together
16=a1+3d
31=a1+6d
Subtract equation 1 from 2
So we have
15=3d
D=15/3
D=5
Therefore, the common difference is 5
