11类NCERT解决方案-第8章二项式定理–练习8.1

Theorem 1:

(a+b)ⁿ = $\sum_{k=0}^{n}$ ⁿC_k a^n-k b^k 

Here, the coefficients ⁿC_k are known as binomial coefficients.

Theorem 2:

(a–b)ⁿ = $\sum_{k=0}^{n}$ (-1)^{n n}C_k a^n-k b^k

展开练习1到5中的每个表达式。

问题1.(1 – 2x) ⁵

解决方案：

According to theorem 2, we have

a = 1

b = 2x

and, n = 5

So, (1 – 2x)⁵ = ⁵C₀ (1)⁵ – ⁵C₁ (1)⁴ (2x)¹ + ⁵C₂ (1)³ (2x)² – ⁵C₃ (1)² (2x)³ + ⁵C₄ (1)¹ (2x)⁴ – ⁵C₅ (2x)⁵

= 1 – 5 (2x) + 10 (4x)² – 10 (8x³) + 5 (16 x⁴) – (32 x⁵)

= 1 – 10x + 40x² – 80x³ + 80x⁴– 32x⁵ 

问题2。 $(\frac{2}{x} - \frac{x}{2})^5$

解决方案：

According to theorem 2, we have

a = $\frac{2}{x}$

b = $\frac{x}{2}$

and, n = 5

So,$(\frac{2}{x} - \frac{x}{2})^5$ = ⁵C₀ ($\frac{2}{x}$)⁵ – ⁵C₁ ($\frac{2}{x}$)⁴ ($\frac{x}{2}$)¹ + ⁵C₂ ($\frac{2}{x}$)³ ($\frac{x}{2}$)² – ⁵C₃ ($\frac{2}{x}$)² ($\frac{x}{2}$)³ + ⁵C₄ ($\frac{2}{x}$)¹ ($\frac{x}{2}$)⁴ – ⁵C₅ ($\frac{x}{2}$)⁵

= $\frac{32}{x^5}$ – 5 $(\frac{16}{x^4}) (\frac{x}{2})$ + 10 $(\frac{8}{x^3}) (\frac{x^2}{4})$ – 10 $(\frac{4}{x^2})$ + 5 $(\frac{2}{x}) (\frac{x^4}{16})$ – $\frac{x^5}{32}$

= $\frac{32}{x^5} - \frac{40}{x^3} + \frac{20}{x} - 5x + \frac{5x^3}{8} - \frac{x^5}{32}$

问题3.(2x – 3) ⁶

解决方案：

According to theorem 2, we have

a = 2x

b = 3

and, n = 6

So, (2x – 3)⁶ = ⁶C₀ (2x)⁶ – ⁶C₁ (2x)⁵ (3)¹ + ⁶C₂ (2x)⁴ (3)² – ⁶C₃ (2x)³ (3)³ + ⁶C₄ (2x)² (3)⁴ – ⁶C₅ (2x)¹ (3)⁵ + ⁶C₆ (3)⁶

= 64x⁶ – 6(32x⁵)(3) + 15 (16x⁴) (9) – 20 (8x³) (27) + 15 (4x²) (81) – 6 (2x) (243) + 729

= 64x⁶ – 576x⁵ + 2160x⁴ – 4320x³ + 4860x² – 2916x + 729

问题4。 $(\frac{x}{3} + \frac{1}{x})^5$

解决方案：

According to theorem 1, we have

a = $\frac{x}{3}$

b = $\frac{1}{x}$

and, n = 5

So, $(\frac{x}{3} + \frac{1}{x})^5$ = ⁵C₀ ($\frac{x}{3}$)⁵ + ⁵C₁ ($\frac{x}{3}$)⁴ ($\frac{1}{x}$)¹ + ⁵C₂ ($\frac{x}{3}$)³ ($\frac{1}{x}$)² + ⁵C₃ ($\frac{x}{3}$)² ($\frac{1}{x}$)³ + ⁵C₄ ($\frac{x}{3}$)¹ ($\frac{1}{x}$)⁴ + ⁵C₅ ($\frac{1}{x}$)⁵

= $\frac{x^5}{243} + 5 (\frac{x^4}{81}) (\frac{1}{x}) + 10 (\frac{x^3}{27}) (\frac{1}{x^2}) + 10 (\frac{x^2}{9}) (\frac{1}{x^3}) + 5 (\frac{x}{3}) (\frac{1}{x^4}) + \frac{243}{x^5}$

= $\frac{x^5}{243} + \frac{5x^3}{81} + \frac{10x}{27} + \frac{10}{9x} + \frac{5}{3x^3} + \frac{1}{x^5}$

问题5 $(x + \frac{1}{x})^6$

解决方案：

According to theorem 1, we have

a = x

b = $\frac{1}{x}$

and, n = 6

So, $(x + \frac{1}{x}) ^ 6$ = ⁶C₀ (x)⁶ + ⁶C₁ (x)⁵ ($\frac{1}{x}$)¹ + ⁶C₂ (x)⁴ ($\frac{1}{x}$)² + ⁶C₃ (x)³ ($\frac{1}{x}$)³ + ⁶C₄ (x)² ($\frac{1}{x}$)⁴ + ⁶C₅ (x)¹ ($\frac{1}{x}$)⁵ + ⁶C₆ ($\frac{1}{x}$)⁶

= $x^6 + 6(x^5)(\frac{1}{x}) + 15 (x^4) (\frac{1}{x^2}) + 20 (x^3) (\frac{1}{x^3}) + 15 (x^2) (\frac{1}{x^4}) + 6 (x) (\frac{1}{x^5}) + (\frac{1}{x^6})$

= $x^6 + 6x^4 + 15x^2 + 20 + \frac{15}{x^2} + \frac{6}{x^4} + \frac{1}{x^6}$

使用二项式定理，评估以下各项：

问题6.(96) ³

解决方案：

Given: (96)³

Here, 96 can be expressed as (100 – 4).

So, (96)³ = (100 – 4)³

According to Theorem 2, we have

= ³C₀ (100)³ – ³C₁ (100)² (4) – ³C₂ (100) (4)²– ³C₃ (4)³

= (100)³ – 3 (100)² (4) + 3 (100) (4)² – (4)³

= 1000000 – 120000 + 4800 – 64

= 884736

问题7.(102) ⁵

解决方案：

Given: (102)⁵

Here, 102 can be expressed as (100 + 2).

So, here (102)⁵ = (100 + 2)⁵

According to Theorem 1, we have

= ⁵C₀ (100)⁵ + ⁵C₁ (100)⁴ (2) + ⁵C₂ (100)³ (2)2 + ⁵C₃ (100)² (2)3 + ⁵C₄ (100) (2)⁴ + ⁵C₅ (2)⁵

= (100)⁵ + 5 (100)⁴ (2) + 10 (100)³ (2)² + 10 (100) (2)³ + 5 (100) (2)⁴ + (2)⁵

= 10000000000 + 1000000000 + 40000000 + 80000 + 8000 + 32

= 11040808032

问题8.(101) ⁴

解决方案：

Given: (101)⁴

Here, 101 can be expressed as (100 + 1).

So, here (101)⁴ = (100 + 1)⁴

According to Theorem 1, we have

= ⁴C₀ (100)⁴ + ⁴C₁ (100)³ (1) + ⁴C₂ (100)² (1)² + ⁴C₃ (100) (1)² + ⁴C₄ (1)⁴

= (100)⁴ + 4 (100)³ + 6 (100)² + 4 (100) + (1)⁴

= 100000000 + 400000 + 60000 + 400 + 1

= 1040604001

问题9.(99) ⁵

解决方案：

Given: (99)⁵

Here, 99 can be expressed as (100 – 1).

So, here (99)⁵ = (100 – 1)⁵

According to Theorem 2, we have

= ⁵C₀ (100)⁵ – ⁵C₁ (100)⁴ (1) + ⁵C₂ (100)³ (1)² – ⁵C₃ (100)² (1)³ + ⁵C₄ (100) (1)⁴ – ⁵C₅ (1)⁵

= (100)⁵ – 5 (100)⁴ + 10 (100)³ – 10 (100)² + 5 (100) – 1

= 1000000000 – 5000000000 + 10000000 – 100000 + 500 – 1

= 9509900499

问题10.使用二项式定理，指出哪个数字更大(1.1) ¹⁰⁰⁰⁰或1000。

解决方案：

Given: (1.1)¹⁰⁰⁰⁰

Here, 1.1 can be expressed as (1 + 0.1)

So, here (1.1)¹⁰⁰⁰⁰ = (1 + 0.1)¹⁰⁰⁰⁰

According to Theorem 1, we have

(1 + 0.1)¹⁰⁰⁰⁰ = ¹⁰⁰⁰⁰C₀ (1)¹⁰⁰⁰⁰ + ¹⁰⁰⁰⁰C₁ (1)⁹⁹⁹⁹ (0.1)¹ + other positive terms

= 1 + 1000 + other positive terms

= 1100 + other positive terms

So, 1100 + other positive terms > 1000

Hence, proved (1.1)¹⁰⁰⁰⁰ > 1000

问题11。找到(a + b) ⁴ –(a – b) ⁴ 。因此，评估(√3+√2) ⁴ –(√3–√2) ⁴ 。

解决方案：

According to Theorem 1, we have

(a + b)⁴ = ⁴C₀ a⁴ + ⁴C₁ a³ b + ⁴C₂ a² b² + ⁴C₃ a b³ + ⁴C₄ b⁴

According to Theorem 2, we have

(a – b)⁴ = ⁴C₀ a⁴ – ⁴C₁ a³ b + ⁴C₂ a² b² – ⁴C₃ a b³ + ⁴C₄ b⁴

Now, (a + b)⁴ – (a – b)⁴ 

= ⁴C₀ a⁴ + ⁴C₁ a³ b + ⁴C₂ a² b² + ⁴C₃ a b³ + ⁴C₄ b⁴ – [⁴C₀ a⁴ – ⁴C₁ a³ b + ⁴C₂ a² b² – ⁴C₃ a b³ + ⁴C₄ b⁴]

= 2 (⁴C₁ a³ b + ⁴C₃ a b³)

= 2 (4a³ b + 4ab³)

= 8ab (a² + b²)                    -(1)

Now, according to Equation(1), we get

a = √3 and b = √2

So, (√3 + √2)⁴ – (√3 – √2)⁴ 

= 8 × √3 × √2 ((√3)² + (√2)²)

= 8 (√6)(3 + 2)

= 40 √6

问题12。找到(x +1) ⁶ +(x-1) ⁶ 。因此，以其他方式评估(√2+1) ⁶ +(√2– 1) ⁶ 。

解决方案：

According to Theorem 1, , we have

(x + 1)⁶ = ⁶C₀ x⁶ + ⁶C₁ x⁵ + ⁶C₂ x⁴ + ⁶C₃ x³ + ⁶C₄ x² + ⁶C₅ x + ⁶C₆

According to Theorem 2, , we have

(x – 1)⁶ = ⁶C₀ x⁶ – ⁶C₁ x⁵ + ⁶C₂ x⁴ – ⁶C₃ x³ + ⁶C₄ x² – ⁶C₅ x + ⁶C₆

Now, (x + 1)⁶ – (x – 1)⁶ 

= ⁶C₀ x⁶ + ⁶C₁ x⁵ + ⁶C₂ x⁴ + ⁶C₃ x³ + ⁶C₄ x² + ⁶C₅ x + ⁶C₆ – [⁶C₀ x⁶ – ⁶C₁ x⁵ + ⁶C₂ x⁴ – ⁶C₃ x³ + ⁶C₄ x₂ – ⁶C₅ x + ⁶C₆]

= 2 [⁶C₀ x⁶ + ⁶C₂ x⁴ + ⁶C₄ x² + ⁶C₆]

= 2 [x⁶ + 15x⁴ + 15x² + 1]                           -(1)

Now, According to Equation(1),

x = √2

So, (√2 + 1)⁶ – (√2 – 1)⁶ 

= 2 [(√2)⁶ + 15(√2)⁴ + 15(√2)² + 1]

= 2 (8 + (15 × 4) + (15 × 2) + 1)

= 2 (8 + 60 + 30 + 1)

= 2 (99)

= 198

问题13：证明只要n是一个正整数，就可以^{将9 n + 1 – 8n – 9整除为64。}

解决方案：

To Prove: 9ⁿ⁺¹ – 8n – 9 = 64 k, where k is some natural number

According to Theorem 1, we have

For a = 1, b = 8 and m = n + 1 we get,

(1 + 8)ⁿ⁺¹ = n + ¹C₀ + n + ¹C₁ (8) + n + ¹C₂ (8)² + …. +  ⁿ⁺¹ C _n+1 (8)ⁿ⁺¹

9ⁿ⁺¹ = 1 + (n + 1) 8 + 8² [ⁿ⁺¹C₂ + ⁿ⁺¹C₃ (8) + …. + ⁿ⁺¹ C _n+1 (8)ⁿ⁺¹]

9ⁿ⁺¹ = 9 + 8n + 64 [ⁿ⁺¹C₂ + ⁿ⁺¹C₃ (8) + …. + ⁿ⁺¹ C _n+1 (8)ⁿ⁺¹]

9ⁿ⁺¹ – 8n – 9 = 64 k

Where k, will be a natural number

Hence, proved 9ⁿ⁺¹ – 8n – 9 is divisible by 64, whenever n is positive integer.

问题14：证明$\sum_{r=0}^{n}$ 3 ^{r n} C _r = 4 ⁿ

解决方案：

As, we know that According to Binomial Theorem,

$\sum_{k=0}^{n}$ ⁿC_k a^n-k b^k = (a + b)ⁿ

By comparing Theorem 1 with question, we get

$\sum_{r=0}^{n}$ 3^{r n}C_r = 4ⁿ

a + b = 4, k = r and b = 3

a = 1.

So, $\sum_{r=0}^{n}$ ⁿC_r a^n-r b^r = (a+b)ⁿ

$\sum_{r=0}^{n}$ ⁿC_r 1^n-r 3^r = (1+3)ⁿ

$\sum_{r=0}^{n}$ ⁿC_r (1) 3^r = 4ⁿ

$\sum_{r=0}^{n}$ ⁿC_r 3_r = 4ⁿ

Hence, Proved