第11类RD Sharma解决方案–第19章算术级数-练习19.1

问题1.如果序列的^第n个项_n由n = n ² – n +1给出，则记下其前五个项。

解决方案：

We have, a_n = n² – n + 1   —(1)

Putting value n = 1 in equation (1), we get

a₁ = (1)² – 1 + 1 = 1

Putting value n = 2 in equation (1), we get

a₂ = (2)² – 2 + 1 = 3

Putting value n = 3 in equation (1), we get

a₃ = (3)² – 3 + 1 = 7

Putting value n = 4 in equation (1), we get

a₄ = (4)² – 4 + 1 = 13

Putting value n = 5 in equation (1), we get

a₅ = (5)² – 5 + 1 = 21

Hence, the five terms of the given n^th term is 1, 3, 7, 13, 21.

问题2。一个序列由_n = n ³ – 6n ² + 11n – 6定义，n∈N。证明该序列的前三个项为零，所有其他项为正。

解决方案：

Given, a_n = n³ – 6n² + 11n – 6  —(1)

Since n∈ N , therefore first three terms are :

a₁ = (1)³ – 6*(1)² + 11*(1) – 6 = 0

a₂ = (2)³ – 6*(2)² + 11*2 – 6 = 0

a₃ = (3)³ – 6*(3)² +11*3 – 6 = 0

Hence, the first three terms a1,a2,a3 are zero.

Equation 1 can be rearranged as:

a_n = (n-2)³ – (n-2)    for n >= 4, an > 0

Hence, all terms excluding first, second and third are positive.

问题3.对于所有n> 1，_找出由1 = 3和_n = 3a _n-1 + 2定义的序列的前四个项。

解决方案：

We have, a_n = 3a_n-1 + 2 and a₁ = 3

Now, 

a₂ = 3*a₁ + 2 = 3*3 + 2 = 11 

a₃ = 3*a₂ + 2 = 3*11 + 2 = 35

a₄ = 3*a₃ + 2 = 3*35 + 2 = 107

Hence, the first four terms are 3, 11, 35, 107.

问题4.在以下每个序列中写下前五个术语：

(i)a ₁ = 1，a _n = a _n-1 + 2，n> 1

解决方案：

We have, a₁ = 1 and a_n = a_n-1 + 2

Now, 

a₂ = a₁ + 2 = 1 + 2 = 3

a₃ = a₂ + 2 = 3 + 2 = 5

a₄ = a₃ + 2 = 5 + 2 = 7

a₅ = a₄ + 2 = 7 + 2 = 9

Hence, the first five terms are 1, 3, 5, 7, 9. 

(ii)a ₁ = 1 = a ₂ ，a _n = a _n-1 + a _n-2 ，n> 2

解决方案：

We have, a₁ = 1, a₂ = 1 and a_n = a_n-1 + a_n-2 

Therefore, 

a₃ = a₂ + a₁ = 2

a₄ = a₃ + a₂ = 3

a₅ = a₄ + a₃ = 5

Hence, the first five terms are 1, 1, 2, 3, 5.

(iii)a ₁ = a ₂ = 2，a _n = a _n-1 – 1 n> 2

解决方案：

We have, 

a₁ = a₂ = 2 and a_n = a_n-1 – 1

Now,

a₃ = a₂ – 1 = 1

a₄ = a₃ – 1 = 0

a₅ = a₄ – 1 = -1

Hence, the first five terms are 2, 2, 1, 0, -1.

问题5.斐波那契数列由_{n = 2的1} = 1 = a 2，a n = a _n-1 + a _n-2定义。对于n = 1、2、3，_{求n + 1} / a _n ，4、5。

解决方案：

We have,
a₁ = a₂ = 1, and

a_n = a_n-1 + a_n-2  for n > 2

Now,

a₃ = a₂ + a₁ = 1 + 1 = 2

a₄ = a₃ + a₂ = 2 + 1 = 3

a₅ = a₄ + a₃ = 3 + 2 = 5

a₆ = a₅ + a₄ = 5 + 3 = 8

Therefore,

for n = 1,    a_n+1/a_n = a₂ / a₁ = 1/1 = 1

for n = 2,    a_n+1/a_n = a₃ / a₂ = 2/1 = 2

for n = 3,    a_n+1/a_n = a₄ / a₃ = 3/2

for n = 4,    a_n+1/a_n = a₅/a₄ = 5/3 

for n = 5,    a_n+1/a_n = a₆/a₅ = 8/5

Hence, 1, 2, 3/2, 5/3, 8/5 are the values for n = 1, 2, 3, 4, 5 respectively.

问题6：证明以下每个序列都是AP。此外，找到共同的区别，并在每种情况下再写3个术语。

(i)3，-1，-5，-9，…

解决方案：

We have, 
a₁ = 3, a₂ = -1, a₃ = -5, a₄ = -9 

Since,

a₂ – a₁ = a₃ – a₂ = a₄ – a₃  = -4

Therefore, It is an A.P with common difference d = -4.

The other three terms are as follows:

a₅ = -9 + -4 = -13

a₆ = -13 + -4 = -17

a₇ = -17 + -4 = -21

(ii)-1、1 / 4、3 / 2、11 / 4，…

解决方案：

We have, 
a₁ = -1, a₂ = 1/4, a₃ = 3/2, a₄ = 11/4 

Since,

 a₂ – a₁ = a₃ – a₂ = a₄ – a₃  = 5/4

Therefore, It is an A.P with common difference d = 5/4.

The other three terms are as follows:

a₅ = 11/4 + 5/4 = 16/4 = 4

a₆ = 16/4 + 5/4 = 21/4

a₇ = 21/4 + 5/4 = 26/4 = 13/2

(iii)√2、3√2、5√2、7√2，…

解决方案：

We have, 
a₁ = √2, a₂ = 3√2, a₃ = 5√2, a₄ = 7√2 

Since,

a₂ – a₁ = a₃ – a₂ = a₄ – a₃  = 2√2

Therefore, It is an A.P with common difference d = 2√2.

The other three terms are as follows:

a₅ = 7√2 + 2√2 = 9√2

a₆ = 9√2 + 2√2 = 11√2

a₇ = 11√2 + 2√2 = 13√2

(iv)9、7、5、3，…

解决方案：

We have, 
a₁ = 9, a₂ = 7, a₃ = 5, a₄ = 3 

Since,  

a₂ – a₁ = a₃ – a₂ = a₄ – a₃  = -2 

Therefore, It is an A.P with common difference d = -2.

The other three terms are as follows:

a₅ = 3 + -2 = 1

a₆ = 1 + -2 = -1

a₇ = -1 + -2 = -3

问题7.序列的^第n个_项由n = 2n + 7给出。证明它是AP。此外，找到其第7个项。

解决方案：

We have, a_n = 2n + 7

Now, 

a₁ = 2 + 7 = 9

a₂ = 4 + 7 = 11

a₃ = 6 + 7 = 13

Since,

a₃ – a₂  =  a₂ – a₁ = 2

The common difference d = 2, Therefore it is an A.P.

Thus, the 7th term is given by:

 a₇ = 2*7 + 7 = 21.

问题8.序列的^第n个_项由n = 2n ² + n + 1给出。表明它不是AP

解决方案：

We have, a_n = 2n² + n + 1 

Now, 

a₁ = 2*(1)² + 1 + 1 = 4

a₂ = 2*(2)² + 2 + 1 = 11

a₃ = 2*(3)² + 3 + 1 = 22

Since,

a₃ – a₂ ≠ a₂ – a₁ 

Therefore, it is not an A.P