Question -
Answer -
Let P (n): 1.3 + 2.4 + 3.5 + … + n. (n+2) = 1/6 n (n+1)(2n+7)
Let us check for n = 1,
P (1): 1.3 = 1/6 × 1 × 2 × 9
: 3 = 3
P (n) is true for n = 1.
Now, let us check for P (n) is true for n = k, and have toprove that P (k + 1) is true.
P (k): 1.3 + 2.4 + 3.5 + … + k. (k+2) = 1/6 k (k+1) (2k+7) …(i)
So,
1.3 + 2.4 + 3.5 + … + k. (k+2) + (k+1) (k+3)
Now, substituting the value of P (k) we get,
= 1/6 k (k+1) (2k+7) + (k+1) (k+3) by using equation (i)
= (k+1) [{k(2k+7)/6} + {(k+3)/1}]
= (k+1) [(2k2 + 7k + 6k + 18)] / 6
= (k+1) [2k2 + 13k + 18] / 6
= (k+1) [2k2 + 9k + 4k + 18] / 6
= (k+1) [2k(k+2) + 9(k+2)] / 6
= (k+1) [(2k+9) (k+2)] / 6
= 1/6 (k+1) (k+2) (2k+9)
P (n) is true for n = k + 1
Hence, P (n) is true for all n ∈ N.