Prove that by math induction 1+2+3+...+n<{(2n+1)^2}/8?

Question

Prove that by math induction 1+2+3+...+n<{(2n+1)^2}/8?

Answer 1

1+2+3+...+n&lt;{(2n+1)^2}/8?

n=1 =&gt; 1&lt;[(2*1+1)^2]/8

1&lt;9/8(true)

n=2 =&gt; 1+2&lt;[(2*2+1)^2]/8

3&lt;25/8(true)

Supose that 1+2+..+k&lt;[(2k+1)^2]/8

and try th demonstrate that

1+2+..+k+(k+1)&lt;[(2(k+1)+1)^2]/8

1+2+..+k+(k+1)&lt;[(2k+1)^2]/8+(k+1)

[(2k+1)^2]/8+(k+1)=[4k^2+4k+1+8k+8]/8=

=[4k^2+12k+9]/8=

=[(2k+3)^2]/8=

=[(2(k+1)+1)^2]/8

QED

--Anna

Answer 2

Let k be a value of n

Pk : 1 + 2 + 3 + -------k &lt;  [ (2k + 1)² ]  / 8

The strategy is then to prove that P1 is true and P (k + 1) is true.

Consider P1

-----------------

LHS = 1

RHS = 9/8

Thus true for P1

Consider P (k + 1)

----------------------

Have to PROVE that :-

1 + 2 + 3 + ----(k + 1) &lt; (2k + 3)² / 8

Now

1 + 2 + 3 + -------k &lt;  [ (2k + 1)² ]  / 8 &lt;------IS TRUE

Thus

1 + 2 + 3 + --(k + 1) &lt; [(2k + 1)²] / 8 + (k + 1)--IS True

RHS

(2k + 1)² + 8k + 8

-----------------------

8

4k² + 12k + 9

------------------

8

Have now to show that

4k² + 5k + 2 &lt; 4k² + 12k + 9

5k + 2 &lt; 12k + 9

THIS IS TRUE for k &gt; 1

Thus P1 true and P k + 1 true

True for all k

Pn is therefore true.

I never find these easy but I think method is OK

Answer 3

Claim:  1 + 2 + 3 + ... + n &lt; (2n + 1)^2 / 8

for n &gt;= 1.

Proof ( by induction )

Base Case:  Let n = 1.  Then

LHS = 1 ... 1

= 1

RHS = (2*1 + 1)^2 / 8

= (2 + 1)^2 / 8

= 3^2 / 8

= 9/8

1 &lt; 9/8, so the formula holds true for n = 1.

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

1 + 2 + 3 + ... + k &lt; (2k + 1)^2 / 8

( we want to prove that the formula holds true for n = k + 1; that is, we want to prove that

1 + 2 + ... + (k + 1) &lt; ( 2(k + 1) + 1)^2 / 8 )

BUT

1 + 2 + 3 + ... + k + 1 =

[1 + 2 + 3 + ... + k] + (k + 1)

Which, by our induction hypothesis, is less than

&lt; [ (2k + 1)^2 / 8 ] + (k + 1)

Merge into one fraction.

= [ (4k^2 + 4k + 1)/8  +  (k + 1) ]

= [ (4k^2 + 4k + 1)/8  +  8(k + 1)/8 ]

= [ { (4k^2 + 4k + 1) + 8(k + 1 }) ] / 8

= [ 4k^2 + 4k + 1 + 8k + 8 ] / 8

= [ 4k^2 + 12k + 9 ] / 8

= [ (2k + 3)^2 ] / 8

= [ (2(k + 1) + 1)^2 ] / 8

Showing that

1 + 2 + ... + (k + 1) &lt; ( 2(k + 1) + 1)^2 / 8

So the formula holds true for n = k + 1.

Therefore, by the principle of mathematical induction,

1 + 2 + 3 + ... + n &lt; (2n + 1)^2 / 8

For all natural numbers n.