Question

bY PMI prove n(n+1)(2n+1) is divisible by 6

Answer 1

when we put value of "n" . we found that the number are devide by 6

hear
n = 1, 2, 3,..............

Answer 2

Answer with Step-by-step explanation:

Let P(n) = n(n+1)(2n+1) is divisible by 6

For  n = 1

∴ $P(1) = 1(1+1)(2\times 1+1)= 6$

which is divisible by 6 therefore P(n) is true for n= 1

Now we assume that P(n) is true for n = k

∴ $k(k+1)(2k+1) = 6A\\\\\Rightarrow (k+1)= \dfrac{6A}{k(2k+1)}$

Now we have to prove that P(n) is true of n= k+1

$P(k+1)= (k+1)(k+1+1)(2(k+1)+1) = (k+1)(k+2)(2k+3)\\\\\Rightarrow \dfrac{6A}{k(2k+1)}(k+2)(2k+3)= 6 (\dfrac{A}{k(2k+1)}(k+2)(2k+3))$

Therefore P(k+1) is divisible by 6

So by PMI P(n) is true for all n i.e n(n+1)(2n+1) is divisible by 6