Proposition: 1*2*3+2*3*4+3*4*5+...+n(n+1)(n+2) = [n(n+1)(n+2)(n+3)]/4(adsbygoogle = window.adsbygoogle || []).push({});

Step (1): If n=1 then LHS (left hand side) = 6, and RHS = 6

Thus, P1 is true.

Step (2): If Pk is true then

k(k+1)(k+2) = [k(k+1)(k+2)(k+3)]/4

Now,

k(k+1)(k+2) + [k+1]([k+1]+1)([k+1]+2) = [k(k+1)(k+2)(k+3)]/4 + [k+1]([k+1]+1)([k+1]+2)

k(k+1)(k+2) +[k+1]([k+1]+1)([k+1]+2) = [k(k+1)(k+2)(k+3)]/4 + (k+1)(k+2)(k+3)

[k(k+1)(k+2)(k+3) + 4(k+1)(k+2)(k+3)]/4 ---> Common denominator - 4.

Then,

let k+1= A, k+2= B, k+3= C

k(A)(B)(C) + 4(A)(B)(C)/4

A(K+4)B(K+4)C(K+4)

Then,

(K+4)(ABC)

am i on the right track?

Then,

(k+4)(k+1)(k+2)(k+3)/4

Then?????? How do i prove that P(k+1) is true whenever Pk and P1 are true???!

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# Help on Mathematical Induction

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