Question

PLEASE HELP ASAP Use mathematical induction to prove the statement is true for all positive integers n, or show why it is false. 1^2+2^2+3^2+...+n^2=n(n+1)(2n+1)/6

Answer 1

Answer:

see below

Step-by-step explanation:

1^2+2^2+3^2+...+n^2=n(n+1)(2n+1)/6

Step1

Verify it for n=1

1^2= 1(1+1)(2*1+1)/6= 1*2*3/6= 6/6=1 - correct

Step2

Assume it is correct for n=k

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

Step3

Prove it is correct for n= k+1

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

prove the above for k+1

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

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

=1/6((k+1)(2k²+k+6k+6))= 1/6(k+1)(2k²+4k+3k+6))=

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

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

Proved for n= k+1 that:

the sum of squares of (k+1) terms equal to  (k+1)(k+2)(2k+3)/6

Answer 2

Answer:

Step-by-step explanation:

Step 1: Consider P(1) that is n = 1

$1^2 = \frac{1(1+1)(2*1+1)}{6}=\frac{6}{6}=1 \checkmark$

Step 2: Suppose the equation is true up to n. That is

$1^2 + 2^2+3^2+........+n^2 = \dfrac{n(n+1)(2n+1)}{6 }$

Step 3: Prove that the equation is true up to (n+1). That is

$1^2 + 2^2+3^2+........+n^2 + (n+1)^2 = \dfrac{(n+1)(n+2)(2n+3)}{6 }$

The easiest way to prove it is to expend the right hand side and prove that the right hand side = the right hand side of step 2 + (n+1)^2

From step 2, add (n+1)^2 both sides. The left hand side will be the left hand side of step 3, now, the right hand side after adding.

$\dfrac{n(n+1)(2n+1)}{6 }+(n+1)^2 = \dfrac{2n^3+3n^2+n}{6}+\dfrac{6n^2+12n+6}{6}$

$=\dfrac{2n^3+9n^2+13n+6}{6}$

If you expend the right hand-side of the step 3, you will see they are same.

Proof done