Question

a sequence is defined by An =n^3- 6n^2+11n-6. show that the first three terms of the sequence are zero and all other terms are positive??? answer me with a smiling face and I will make you as the brainliest answer.​

Answer 1

Answer:

Step-by-step explanation:

Answer 2

$\large{\underline{\rm{\green{\bf{Question:-}}}}}$

A sequence is defined by an = n³ – 6 n² + 11 n – 6, n ∈ N. Show that the first three terms of the sequence are zero and all other terms are positive.

$\large{\underline{\rm{\green{\bf{Given:-}}}}}$

$\sf a_n=n^{3} -6n^{2}+11n-6, \: n \in N$

$\large{\underline{\rm{\green{\bf{To \: Find:-}}}}}$

The first three terms of the sequence are zero and all other terms are positive.

$\large{\underline{\rm{\green{\bf{Analysis:-}}}}}$

By using the values n = 1, 2, 3 we can find the first three terms.

$\large{\underline{\rm{\green{\bf{Solution:-}}}}}$

Given that, $\sf a_n=n^{3} -6n^{2}+11n-6, \: n \in N$

Using the values n = 1, 2, 3 we can find the first three terms.

When n = 1:

$\sf a_1=(1)^{3}-6(1)^{2}+11(1)-6$

$\sf =1-6+11-6$

$\sf =12-12$

$\sf =0$

When n = 2:

$\sf a_2=(2)^{3}-6(2)^{2}+11(2)-6$

$\sf =8-6(4)+22-6$

$\sf =8-24+22-6$

$\sf =30-30$

$\sf =0$

When n = 3:

$\sf a_3=(3)^{3} -6(3)^{2}+11(3)-6$

$\sf =27-6(9)+33-6$

$\sf =27-54+33-6$

$\sf = 60-60$

$\sf =0$

This shows that the first three terms of the sequence is zero.

Now, checking for when n = n

$\sf a_n = n^{3} - 6n^{2} + 11n - 6$

$\sf = n^{3} - 6n^{2} + 11n - 6 - n + n - 2 + 2$

$\sf =(n)^{3} - 3 \times 2n(n - 2) - (2)3 - n + 2$

By using the formula,

$\sf [(a - b)3 = (a)3 - (b)3 - 3ab(a - b)]$

$\sf =a_n = (n - 2)^{3} - (n - 2)$

Here, n – 2 will always be positive for n > 3

∴ $\sf a_n$ is always positive for n > 3