n(n+1)(n+5) is a multiple of 3

please tell the two identity 1.n=1,n=k+1

Open in App

Solution

p(n)=n(n+1)(n+5)

p(1)=1(1+1)(1+5)=1x2x6=12

let p(k) b true

p(k)=k(k+1)(k+5)

on solving

=k³+6k²+5k=3y( as itsa multiple of 3)

k³=3y-6k²-5k ------(1)

p(k+1)= k+1 ( k+1 +1) k+1 +5)

=k+1 ( k+2) k+6)

on solving =k³+9k²+20k +12

putting value of k³ from (1)

=3y-6k²-5k+9k²+20k +12

=3y+3k²+15k +12

3(y+k²5k +4)

therefore we can say p(k+1) is a multiple of 3

p(k) is true so p(n) is also true and its a multiple of 3

Suggest Corrections

Similar questions

6

n - \frac{n (n - 1) (n - 2)}{⌊ 3} \cdot 3 + \frac{n (n - 1) (n - 2) (n - 3) (n - 4)}{⌊ 5} \cdot 3^{2} - . . . . .,

n - \frac{n (n - 1) (n - 2)}{⌊ 3} \cdot \frac{1}{3} + \frac{n (n - 1) (n - 2) (n - 3) (n - 4)}{⌊ 5} \cdot \frac{1}{3^{2}} - . . . . .,

n

n^{n + 1} - n {(n - 1)}^{n + 1} + \frac{n (n - 1)}{2!} {(n - 2)}^{n + 1} - \dots = \frac{1}{2} n (n + 1)!

n^{n} - (n + 1) {(n - 1)}^{n} + \frac{(n + 1) n}{2!} {(n - 2)}^{n} - \dots = 1

n

1^{n} - n 2^{n} + \frac{n (n - 1)}{1 \cdot 2} 3^{n} - \dots = {(- 1)}^{n} n!

{(n + p)}^{n} - n {(n + p - 1)}^{n} + \frac{n (n - 1)}{2!} {(n + p - 2)}^{n} - \dots = n!

n + 1

$n (n + 1) (n + 5)$ is a multiple of 3 for all $n ϵ N$ .

7^{n} {1 + \frac{n}{7} + \frac{n (n - 1)}{7.14} + \frac{n (n - 1) (n - 2)}{7.14.21} + . . . .}

= 4^{n} {1 + \frac{n}{2} + \frac{n (n - 1)}{2.4} + \frac{n (n - 1) (n - 2)}{2.4.6} + . . . .}

$n^{n} - n (n - 1)^{n} + \frac{n (n - 1)}{2!} (n - 2)^{n} - . . . \infty = \frac{3}{a} \times n!$ .

Find $a$

Join BYJU'S Learning Program