Question

Problem 2
Prove that
1 2 + 2 2 + 3 2 + ... + n 2 = n (n + 1) (2n + 1)/ 6
For all positive integers n.

Answer 1

Answer:Statement P (n) is defined by  

1 2 + 2 2 + 3 2 + ... + n 2 = n (n + 1) (2n + 1)/ 2  

STEP 1: We first show that p (1) is true.  

Left Side = 1 2 = 1  

Right Side = 1 (1 + 1) (2*1 + 1)/ 6 = 1  

Both sides of the statement are equal hence p (1) is true.  

STEP 2: We now assume that p (k) is true  

1 2 + 2 2 + 3 2 + ... + k 2 = k (k + 1) (2k + 1)/ 6  

and show that p (k + 1) is true by adding (k + 1) 2 to both sides of the above statement  

1 2 + 2 2 + 3 2 + ... + k 2 + (k + 1) 2 = k (k + 1) (2k + 1)/ 6 + (k + 1) 2  

Set common denominator and factor k + 1 on the right side  

= (k + 1) [ k (2k + 1)+ 6 (k + 1) ] /6  

Expand k (2k + 1)+ 6 (k + 1)  

= (k + 1) [ 2k 2 + 7k + 6 ] /6  

Now factor 2k 2 + 7k + 6.  

= (k + 1) [ (k + 2) (2k + 3) ] /6  

We have started from the statement P(k) and have shown that  

1 2 + 2 2 + 3 2 + ... + k 2 + (k + 1) 2 = (k + 1) [ (k + 2) (2k + 3) ] /6  

Which is the statement P(k + 1).

Step-by-step explanation:

Answer 2

Answer:

Let Pn be the statement 1^2 + 2^2 + 3^2 + ... +

n^2 = [ n(n + 1)(2n + 1) ] /6

(The Anchor) P1 is true because 1^2 = [ 1(2)(3) ]/6

(The Inductive hypothesis) Assume that Pk is true, So that

1^2 + 2^2 + ...... + k^2 = k (k + 1) (2k + 1) /6

(The Inductive Step) The next term on the left hand side would be (K + 1) ^2 . We add this to both sides of Pk and get.

$1 {}^{2} + 2 {}^{2} + ... + k {}^{2} + (k + 1) {}^{2} = k(k + 1)(2k + 1) \div 6 + (k + 1) {}^{2}$

$= k(k + 1)(2k + 1) + 6(k + 1) {}^{2} \div 6$

$= (k + 1)(2k {}^{2} + k + 6k + 6) \div 6$

$= (k + 1)(k + 2)(2k + 3) \div 6$

$= (k + 1)( (k + 1) + 1)(2(k + 1) + 1) \div 6$

This is exactly the statement P k + 1,So the equation is true for n = k + 1 . Therefore Pn is true for all positive integers, by mathematical induction.

Step-by-step explanation:

@SSR