Let T(n) denote the given statement
1. 4. 7 + 2. 5. 8 + 3. 6. 9 + ... + n (n + 3) (n + 6) = \(\frac{n}{4}\) (n + 1) (n + 6) (n + 7)
Basic Step:
For n = 1, LHS = 1. 4. 7 = 28
RHS = \(\frac{1}{4}\) (1 + 1) (1 + 6) (1 + 7) = 28
⇒ LHS = RHS ⇒ T(1) is true
Induction Step:
Assume T(k) is true for all k∈N, i.e.,
1. 4. 7 + 2. 5. 8 + 3. 6. 9 + ... + k (k + 3) (k + 6) =\(\frac{k}{4}\)(k + 1) (k + 6) (k + 7)
Now we shall show that T (k + 1) is also true.
To obtain T (k + 1) add the (k + 1)th term, i.e., (k + 1) (k + 1 + 3) (k + 1 + 6) = (k + 1) (k + 4) (k + 7) to both the sides of T(k). Then,
1. 4. 7 + 2. 5. 8 + 3. 6. 9 + ... + k (k + 3) (k + 6) + (k + 1) (k + 4) (k + 7)
= \(\frac{k}{4}\) (k + 1) (k + 6) (k + 7) + (k + 1) (k + 4) (k + 7) = (k + 1) (k + 7)\(\big[\frac{k}{6}(k+6)+(k+4)\big]\)
\(\frac{(k+1)(k+7)}{4}\) [k2 + 6k + 4k + 16] = \(\frac{(k+1)}{4}\) (k + 7) (k2+10k +16)
\(\frac{(k+1)}{4}\)(k + 7)(k + 2)(k + 8) = \(\frac{1}{4}\)(k + 1)(k + 2)(k + 7)(k + 8)
⇒ T (k + 1) is true, assuming T (k) is true.
⇒ T(n) is true for all n∈N