特殊系列–序列和序列| 11年级数学

序列可以定义为给定序列的所有数字之和。序列是无限的。同样，级数也可以是有限的或无限的。例如，考虑一个序列为1、3、5、7…，那么这些项的序列将为1 + 3 + 5 + 7 +…。 。以某种方式特殊的系列称为特殊系列。以下是特殊系列的三种类型。

1. 1 + 2 + 3 +…+ n(前n个自然数之和)
2. 1 ² + 2 ² + 3 ² +…+ n ² (前n个自然数的平方和)
3. 1 ³ + 2 ³ + 3 ³ +…+ n ³ (前n个自然数的立方的总和)

在本文中，我们将看到如何获得所有这些系列的公式。

特殊系列1：前n个自然数之和

该系列的结果如下：

1+ 2 + 3 + 4 + …. + n = n (n + 1) / 2

证明：

Let S_n = 1 + 2 + 3 + 4 + … + n

We can see that this is an Arithmetic Progression with the first term (a) = 1 and common difference (d) =1 and there are n term 

So, Sum of n terms = n/2 (2 x a + (n – 1) x d)

Putting the values for this series we will get 

S_n = n/2(2 x 1 + (n – 1) x 1)

S_n = n/2(2 + n – 1)

S_n = n(n + 1)/2

Hence Proved.

例子

问题。找到以下系列的总和3 + 4 + 5 — + 25?

解决方案：

Let S_n = 3+ 4 + 5 — + 25

Now we can also write it like this 

S_n + 1 + 2 = 1 + 2 + 3 + 4 —- + 25

Clearly now it is the sum of first 25 natural number we can be written like this 

S_n + 1+ 2 = 25 (25 + 1) / 2

S_n = 325 – 1 – 2

S_n = 322 

特殊系列2：前n个自然数的平方和

该系列的结果如下：

1² + 2² + 3² +… + n²  = n(n + 1) (2n + 1)/6

证明：

Let S_n = 1² + 2² + 3² +… + n²   —eq 1

We know that, k³ – (k – 1)³ = 3k² – 3k + 1  — eq 2

We know that, (a – b)³ = a³ – b³ – 3a²b + 3ab²

So, k³ – (k – 1)³

= k³ – k³ +1 + 3k² – 3k

= 3k² – 3k +1

Putting k = 1, 2…, n successively in eq 2, we obtain

1³ – 0³ = 3(1)² – 3(1) + 1

2³ – 1³ = 3(2)² – 3(2) + 1

3³ – 2³ = 3(3)² – 3(3) + 1

…………………………………

…………………………………

………………………………..

n³ – (n – 1)³= 3(n)² – 3(n) + 1

Adding both sides of all above equations, we get

n³ – 0³ = 3 (1² + 2² + 3² + … + n²)  – 3 (1 + 2 + 3 + … + n) + n

We can write this like:

n³ = 3 ∑(k²) – 3∑(k) +n, where 1 ≤ k ≤ n  — eq(3)

We know that,

 ∑(k)  (where 1 ≤ k ≤ n ) = 1 + 2 + 3 + 4 — n = n(n + 1)/2  —eq(4)

and eq 1  can also be written like this

S_n = ∑(k²), where 1 ≤ k ≤ n  — eq(1)

Now, putting these values in eq 3

n³ = 3S_n – 3(n)(n + 1)/ 2 + n

n³ + 3 (n) (n + 1)/2 – n = 3S_n

(2n³ + 3n² + 3n – 2n)/2 = 3S_n

(2n³ + 3n² + n)/6 = S_n

n(2n² + 3n + 1)/6 = S_n

n(2n² + n + 2n + 1)/6 = S_n

n(n(2n + 1) + 1(2n + 1))/6 = S_n

n(n + 1)(2n + 1)/6 = S_n

S_n = n(n + 1)(2n + 1)/6

Hence proved.

例子

问题1.找到第n个项为n ² + n + 1的级数的n个项之和?

解决方案：

Given that , 

a_n = n² + n + 1

Thus, the sum to n terms is given by

S_n = ∑ak (where 1 ≤ k ≤ n ) = ∑ k² +  ∑ k +  ∑1  (where 1 ≤ k ≤ n)

= n(n + 1) (2n + 1)/6 + n (n + 1)/2 + n

= (n(n + 1) (2n + 1) + 3n(n + 1) + 6n)/6

= ((n²+ n) (2n + 1) + 3n² + 3n + 6n)/6

= (2n³ + 2n² + n² + n + 3n² + 9n)/6

= (2n³ + 6n² + 10n)/6

问题2。找到最多n个项的以下系列的总和1 +1 + 2 +1 +1 2 2 + 3 +1 + 2 +1 3 + 4 +-?

解决方案：

If we observe the  series carefully we can write it like this 

S_n =(1) + (1 + 2) + (1 + 2 + 3) + (1 + 2 + 3 + 4) + ——

We can say that we have to find sum of the sum of  first n natural number.

So we can write S_n=  Σ((i(i + 1))/2), where 1 ≤ i ≤ n

= (1/2)Σ (i(i + 1))

= (1/2)Σ (i² + i)

= (1/2)(Σ i² + Σ i)

We know Σ i² = n (n + 1) (2n + 1) / 6 and  

Σ i = n (n + 1) / 2.

Substituting the value, we get,

Sum = (1/2)((n(n + 1)(2n + 1) / 6) + (n( n + 1) / 2))  

        = n(n + 1)/2 [(2n + 1)/6 + 1/2]

        = n(n + 1)(n + 2) / 6

特殊系列3：前n个自然数的立方和

该系列的结果如下：

1³ + 2³ + 3³ + … + n³  = (n (n + 1)/2)²

证明：

Let S_n = 1³ + 2³ + 3³ +… + n³   —eq 1

We know that, (k + 1)⁴ – (k)⁴ = 4k³ + 6k² + 4k + 1              — eq 2

We know that, (a+b)⁴ = (a² +b² +2ab)²

= a⁴ + b⁴ + 6a²b² + 4a³b + 4ab³

So, (k + 1)⁴ – (k)⁴

= k⁴ + 1 + 6k² + 4k³ + 4k- k⁴ 

= 4k³ +6k² + 4k +1

Putting k = 1, 2…, n successively in eq 2 , we obtain

(1 + 1)⁴ – 1⁴ = 4(1)³ + 6(1)² +  4(1) + 1

(2 + 1)⁴ – 2⁴ = 4(2)³ + 6(2)² + 4(2) + 1

…………………………………

…………………………………

………………………………..

(n + 1)⁴ – (n)⁴ = 4(n)³ + 6n² + 4n + 1

Adding both sides of all the above equations, we get

(n + 1)⁴ – 1⁴ = 4 (1³ + 2³+ 3³ + … + n³) + 6(1² + 2²+ 3² + 4² + 5²)  + 4 (1 + 2 + 3 + … + n) + n

We can write this like:

(n + 1)⁴ – 1⁴ = 4 ∑ (k³) + 6∑(k²) + 4∑(k) + n    where 1 ≤ k ≤ n  — eq(3)

We know that ,

∑(k)  (where 1 ≤ k ≤ n ) = 1 + 2 + 3 + 4 — n = n (n + 1)/2  —eq(4)

∑(k²)  (where 1 ≤ k ≤ n ) = 1² + 2² + 3² + 4² — n² = n (n + 1) (2n + 1)/6  —eq(5)

and eq 1  can also be written like this

S_n = ∑(k³) , where 1 ≤ k ≤ n  — eq(1)

Now, putting these values in eq 3

(n + 1)⁴ -1⁴ = 4S_n+ 6(n) (n + 1) (2n + 1)/6 + 4 (n) (n + 1)/2 + n

n⁴  + 6n² + 4n³ + 4n – (n)(2n² + 3n + 1) – 2(n)(n + 1) – n = 4S_n

n⁴ + 6n² + 4n³ + 4n – 2n³ – 3n² – n – 2n² – 2n – n = 4S_n

n⁴ + n² + 2n³ = 4S_n 

n² (n² + 1 + 2n) = 4S_n

n² (n + 1)² = 4S_n

S_n = (n(n + 1)/2)²

Hence proved.

例子

问题。找到以下分数的值(1 ³ + 2 ³ + 3 ³ — + 9 ³ )/(1 + 2 + 3 — + 9)?

解决方案：

Sum of first n natural number : n(n + 1)/2

Sum of cube of first n natural number : (n(n + 1)/2)²

So, (1³ + 2³ + 3³ —-+  n³) / (1+ 2+ 3 —- +n)

= ((n(n + 1)/2)²) / (n(n + 1)/2)

= n(n + 1)/2

Now as we can see that value of n is 9 in the question,

= 9 (9 + 1) / 2

= 9 x 5

= 45