induction proof.?

Question

Use induction to prove that

1^2 x 2 + 2^2 x 3 + 3^2 x 4 +…+ n^2 x (n + 1) =( n(n+1)(3n^2 + 7n +2))/ 12

for all positive integers n.

how do u prove it for P(m+1) like LHS =RHS and that.

Answer 1

Assume it's true for terms up to n^2 x (n + 1), i.e. that

1^2 x 2 + 2^2 x 3 + 3^2 x 4 +…+ n^2 x (n + 1)

= (n(n+1)(3n^2 + 7n +2))/12.

In fact, since the last quadratric term itself FACTORS as (n+2)(3n+1), write this whole sum as follows:

1^2 x 2 + 2^2 x 3 + 3^2 x 4 +…+ n^2 x (n + 1)

= n(n+1)(n+2)(3n+1)/12 = S(n), say.

Now consider what results if one adds the next term to the LHS. That term will be (n+1)^2 x (n+2). Adding it on the RHS, one will obtain:

{[(n+1)(n+2)]/12}[n(3n+1)+12(n +1)]

= {[(n+1)(n+2)]/12} [3n^2 +13n +12]

= {[(n+1)(n+2)]/12}[(n+3)(3n +4)],

which is EXACTLY the original form S(n) for the LHS, but with
' n ' replaced by ' n+1 ', i.e. S(n+1).

Thus, if it's true for 'n,' then the expression for this sum is also true for 'n+1.'

But, for n = 1, the LHS is 1^2 x 2 = 2, while S(1) = 1*2*3*4/12 = 2. Good! --- it IS true for n = 1, so the induction can start.

Therefore it's true for all positive integers ' n.'

Live long and prosper.

POSTSCRIPT:

I have to wonder why the original expression S(n) wasn't completely factored, as I showed that it could be, above. I think you'll agree that the complete factoring helped to reduce the work involved in confirming the form for the result by induction.

Answer 2

1^2 x 2 + 2^2 x 3 + ... + n^2 x (n + 1) = ( n(n+1)(3n^2 + 7n +2))/ 12

Proof by induction:

1) Base case: Let n = 1. Then
LHS = 1^2 x 2 ... 1^2 x 2 = 2 ... 2 = 2
RHS = 1(1 + 1)(3(1)^2 + 7(1) + 2))/12
= 1(2)(12)/12
= 2
LHS = RHS, so the formula holds true for n = 1.

2) Induction Hypothesis. Assume the formula holds true for n = k. That is, assume that

1^2 x 2 + 2^2 x 3 + ... + k^2 x (k + 1) = ( k(k+1)(3k^2 + 7k +2))/ 12

(We want to prove that the formula holds true for n = k + 1.)
BUT the sum of k + 1 terms is the same as the sum of k terms, plus the (k + 1)st term. i.e.

1^2 x 2 + 2^2 x 3 + ... + (k + 1)^2 x (k + 1 + 1) =
{1^2 x 2 + 2^2 x 3 + ... + k^2 x (k + 1)} + (k + 1)^2 x (k + 1 + 1)

The terms in { } brackets is precisely our induction hypothesis; replace that with what we assumed was true.

{ ( k(k + 1)(3k^2 + 7k +2))/ 12 } + (k + 1)^2 x (k + 1 + 1)

{ ( k(k + 1)(3k + 1)(k + 2) )/12 } + (k + 1)^2 (k + 2)

Factor (1/12)(k + 1)(k + 2) to obtain

= (1/12)(k + 1)(k + 2) [ k(3k + 1) + 12(k + 1) ]

= (1/12)(k + 1)(k + 2) [ 3k^2 + k + 12k + 12 ]

= (1/12)(k + 1)(k + 2) [ 3k^2 + 13k + 12 ]

= (1/12)(k + 1)(k + 2) [ 3k^2 + 13k + 12 ]

= (1/12)(k + 1)(k + 2) [ 3k^2 + 6k + 3 + 7k + 7 + 2 ]

= (1/12)(k + 1)(k + 2) [ 3(k^2 + 2k + 1) + 7k + 7 + 2 ]

= (1/12)(k + 1)(k + 2) [ 3(k + 1)^2 + 7(k + 1) + 2 ]

= (1/12)(k + 1)(k + 1 + 1) [ 3(k + 1)^2 + 7(k + 1) + 2 ]

= { (k + 1)(k + 1 + 1) [ 3(k + 1)^2 + 7(k + 1) + 2 ] } / 12

Clearly showing that the formula holds true for n = k + 1.

Thus by the principle of mathematical induction,

1^2 x 2 + 2^2 x 3 + 3^2 x 4 +…+ n^2 x (n + 1) =( n(n+1)(3n^2 + 7n +2))/ 12
for all positive integers n.

Answer 3

Prove for P(1), generalise for P(m) and then prove for P(m+1).