Chapter 12.1, Problem 80E

a.

To determine

Prove that the first four terms of the sequence are sw $1,4,9,16,\mathrm{........}$ .

Explanation of Solution

Given:

It is given in the question that the sequence is $a_n=n^2$.

Concept Used:

In this, use the concept of putting the counting value in the sequence and make the answer and prove it.

Proof:

Here, the sequence is $a_n=n^2$ .

For,

  $\begin{array}{l}n=1\Rightarrow 1^2=1\\ n=2\Rightarrow 2^2=4\\ n=3\Rightarrow 3^2=9\\ n=4\Rightarrow 4^2=16\end{array}$

Hence, the sequence of $n^2$ is $1,4,9,16,\mathrm{........}$ .

b.

To determine

Prove that the first four terms of the sequence is $1,4,9,16,\mathrm{........}$ .

Explanation of Solution

Given:

It is given in the question that the sequence is $a_n=n^2+(n-1)(n-2)(n-3)(n-4)$ .

Concept Used:

In this, use the concept of putting the counting value in the sequence and make the answer and prove it.

Proof:

Here, the sequence is $a_n=n^2+(n-1)(n-2)(n-3)(n-4)$ .

For,

  $\begin{array}{l}n=1\Rightarrow 1^2+(1-1)(1-2)(1-3)(1-4)=1+0=1\\ n=2\Rightarrow 2^2+(2-1)(2-2)(2-3)(2-4)=4+0=4\\ n=3\Rightarrow 3^2+(3-1)(3-2)(3-3)(3-4)=9+0=9\\ n=4\Rightarrow 4^2+(4-1)(4-2)(4-3)(4-4)=16+0=16\end{array}$

The sequence of $n^2+(n-1)(n-2)(n-3)(n-4)$ are $1,4,9,16,\mathrm{........}$ .

c.

To determine

Calculate a sequence whose first six terms are the same as those of $a_n=n^2$ but whose succeeding terms differ from this sequence.

Answer to Problem 80E

The sequence which will be same is $b_n=n^2+(n-1)(n-2)(n-3)(n-4)(n-5)(n-6)$ .

Explanation of Solution

Given: It is given in the question that the two sequence are $a_n=n^2$ and $a_n=n^2+(n-1)(n-2)(n-3)(n-4)$ .

Concept Used:

In this, use the concept of taking an sequence $b_n=a_n+candc=0$ (let) and do the question.

Calculation: It is given in the problem that for the first six terms the sequence is $a_n=n^2$ , but after that six terms the sequence differ. So the terms of the sequence can be given by $b_n=a_n+candc=0$ for the first term. Hence, the value of “c” can be $c=(n-1)(n-2)(n-3)(n-4)(n-5)(n-6)$ .

So, the terms will be given by : $b_n=n^2+(n-1)(n-2)(n-3)(n-4)(n-5)(n-6)$

Conclusion:

The sequence is $b_n=n^2+(n-1)(n-2)(n-3)(n-4)(n-5)(n-6)$ .

d.

To determine

Calculate the two different that begins with $2,4,8,16,\mathrm{........}$ .

Answer to Problem 80E

The two sequence are $a_n=2^{n}and{a}_n=2^n+(1-n)(2-n)(3-n)(4-n)$ .

Explanation of Solution

Given:

It is given in the question that the term is $2,4,8,16,\mathrm{........}$ .

Concept Used:

In this, use the concept of taking a sequence $a_n=2^n(let)$ and understand the term and make the question.

Calculation:

Since, the sequence is $2,4,8,16,\mathrm{........}$ .

For two different sequences that begins with above terms are :

  $1.a_n=2^n,$ Here , it will get same sequence by putting $n=1,2,3,4,\mathrm{..........}$

  $2.a_n=2^n+C,$ Here it will get same sequence by putting $n=1,2,3,4,\mathrm{..........}$ .

  $C=(1-n)(2-n)(3-n)(4-n)$

Then, $a_n=2^n+(1-n)(2-n)(3-n)(4-n)$ .

Conclusion:

The two sequence are $a_n=2^{n}and{a}_n=2^n+(1-n)(2-n)(3-n)(4-n)$ .