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Prove by the principle of induction that 

1. 4. 7 + 2. 5. 8 + 3. 6. 9 + ... + n (n + 3) (n + 6)= \(\frac{n}{4}\) (n + 1) (n + 6) (n + 7).

1 Answer

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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

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