Question

plz answer this question fast

Answer 1

HOLA!!

We have to show that the first three terms are zero and all other terms are positive!!

So, just put the values of 1, 2, and 3, in place of n.
You must get zero.
Lets check it:

Given equation:

[tex] n^{3} -6n^{2}+11n-6 [/tex]

On putting 1, we get:

[tex]n^{3} -6n^{2}+11n-6 \\= 1^{3}-6\times(1)^{2} + 11\times1 - 6 \\ = 1 - 6 + 11 - 6 \\= 12 - 12 \\ = 0[/tex]

Thus, the first term is zero (Proved).

On putting 2 for the value of n, we get:

[tex]n^{3} -6n^{2}+11n-6 \\= (2)^{3} - 6\times(2)^{2} + 11\times2 - 6 \\= 8-6\times4+22-6 \\=8-24 + 16 \\=16-16 \\=0[/tex]

Thus, the second term is zero (Proved).

Now, on putting 3, we get:

[tex]n^{3} -6n^{2}+11n-6 \\=(3)^{3} - 6\times(3)^{2} + 11\times3 - 6 \\=27 - 54 + 33 - 6 \\=27-27 \\=0[/tex]

Hence, third term is also zero (Proved).

Similarly, put any other value beyond 3, you will get a positive value greater than zero.

For example, let's try putting 4:

[tex]n^{3} -6n^{2}+11n-6 \\= 4^{3} - 6\times(4)^{2} + 11\times4 - 6 \\=64 - 6\times16 + 44 - 6 \\=64 - 96 + 38 \\= 102 - 96 \\=6[/tex]

6 is a positive value....

[PROVED]

Hope my answer is satisfactory and obliging...
THANKS!

Answer 2

HELLO DEAR,
$put \: n = 1 \: and \: \: 2 \: and \: 3 \\ \\ \\ n = 1 \\ \\ a_{1} = {1}^{3} - 6 {(1)}^{2} + 11 \times 1 - 6 \\ \\ = 1 - 6 + 11 - 6 \\ \\ = - 5 + 5 = 0$
put n=2
$a_{2} = {2}^{3} - 6 {(2)}^{2} + 11(2) - 6 \\ \\ = 8 - 24 + 22 - 6 \\ \\ = - 16 + 16 = 0$
put n=3

$a_{3} = {3}^{3} - 6 {(3)}^{2} + 11(3) - 6 \\ \\ = 27 - 54 + 33 - 6 \\ \\ = - 27 + 27 = 0$
HENCE first three terms are "0"

now the NEXT terms are

put n=4

$a_{4} = {4}^{3} - 6( {4}^{2} ) + 11(4) - 6\\ \\ = 64 - 96+ 44 - 6 \\ \\ = 108 - 102 = 6$

put n=5

$a_{5} = {5}^{3} - 6( {5}^{2} ) + 11(5) - 6 \\ \\ = 125 - 150 + 55 - 6 \\ \\ = - 25 + 49 = 24$

I HOPE ITS HELP YOU DEAR,
THANKS