第 11 类 RD Sharma 解决方案 - 第 18 章二项式定理 - 练习 18.2 |设置 1

问题 1. 求 (2x – 1/x ² ) ²⁵展开式中的第 11 项和第 11 项。

解决方案：

We are given, (2x – 1/x²)²⁵.

The given expression contains 25 + 1 = 26 terms.

So, the 11th term from the end is the (26 − 11 + 1) th term = 16th term from the beginning.

Hence, T₁₆ = T₁₅₊₁ = ²⁵C₁₅ (2x)^25-15 (−1/x²)¹⁵

= ²⁵C₁₅ (2¹⁰) (x)¹⁰ (−1/x³⁰)

= – ²⁵C₁₅ (2¹⁰/ x²⁰)

Now, the 11th term from the beginning is,

T₁₁ = T₁₀₊₁ = ²⁵C₁₀ (2x)^25-10 (−1/x²)¹⁰

= ²⁵C₁₀ (2¹⁵) (x)¹⁵ (1/x²⁰)

= ²⁵C₁₀ (2¹⁵/ x⁵)

问题 2. 求 (3x ² – 1/x ³ ) ¹⁰展开式中的第 7 项。

解决方案：

We are given, (3x² – 1/x³)¹⁰. 

The 7th term of the expression is given by,

T₇ = T₆₊₁

= ¹⁰C₆ (3x²)¹⁰⁻⁶ (−1/x³)⁶

= ¹⁰C₆ (3)⁴ (x)⁸ (1/x¹⁸)

= $\frac{10×9×8×7×81}{4×3×2×x^{10}}$

= 17010/ x¹⁰

问题 3. 在 (3x – 1/x ² ) ¹⁰的展开式中找出倒数第 5 项。

解决方案：

We are given, (3x – 1/x²)¹⁰

The 5th term from the end is the (11 – 5 + 1)th = 7th term from the beginning.

So, T₇ = T₆₊₁

= ¹⁰C₆ (3x)^10-6 (–1/x²)⁶

= ¹⁰C₆ (3)⁴ (x)⁴ (1/x¹²)

= $\frac{10×9×8×7×81}{4×3×2×x^8}$

= 17010/ x⁸

问题 4. 求 (x ^3/2 y ^1/2 – x ^1/2 y ^3/2 ) ¹⁰展开式中的第 8 项。

解决方案：

We are given, (x^3/2 y^1/2 – x^1/2 y^3/2)¹⁰. 

The 8th term of the expression is given by,

T₈ = T₇₊₁

 = ¹⁰C₇ (x^3/2 y^1/2)^10–7 (–x^1/2 y^3/2)⁷

= $\frac{-(10×9×8)}{3×2} x^{9/2} y^{3/2} (x^{7/2} y^{21/2})$

= –120 x⁸y¹²

问题 5. 求 (4x/5 + 5/2x) ⁸展开式中的第 7 项。

解决方案：

We are given, (4x/5 + 5/2x)⁸.

The 8th term of the expression is given by,

T₇ = T₆₊₁

= $^8C_6(\frac{4x}{5})^{8-6}(\frac{5}{2x})^6$

= $\frac{8×7×6!}{6!2!}(\frac{16x^2}{25})(\frac{125×125}{64x^6})$

= 4375/ x⁴

问题 6. 求 (x + 2/x) ⁹的展开式中的第 4 项和第 4 项。

解决方案：

We are given, (x + 2/x)⁹.

The given expression contains 9 + 1 = 10 terms.

So, the 4th term from the end is the (10 − 4 + 1) th term = 7th term from the beginning.

Hence, T₇ = T₆₊₁ = ⁹C₆ (x)^9-6 (2/x)⁶

= $\frac{9×8×7}{3×2}(x^3)(\frac{64}{x^6})$

= 5376/ x³

Now, the 4th term from the beginning is,

T₄ = T₃₊₁ = ⁹C₃ (2x)^9-3 (2/x)³

= $\frac{9×8×7}{3×2}(x^6)(\frac{8}{x^3})$

= 672 x³

问题 7. 找出 (4x/5 – 5/2x) ⁹展开式中倒数第 4 项。

解决方案：

We are given, (4x/5 – 5/2x)⁹

The 4th term from the end is (10 − 4 + 1)th term = 7th term, from the beginning.

T₇ = T₆₊₁

= $^9C_6(\frac{4x}{5})^{9-6}(\frac{5}{2x})^6$

= $\frac{9×8×7}{3×2}(\frac{64x^3}{125})(\frac{125×125}{64x^6})$

= 10500/ x³

问题 8. 在 (2x ² – 3/2x) ⁸的展开式中找出倒数第 7 项。

解决方案：

We are given, (2x² – 3/2x)⁸

The 7th term from the end is (9 − 7 + 1)th term = 3rd term, from the beginning.

T₃ = T₂₊₁

= $^8C_2(2x^2)^{8-2}(\frac{-3}{2x})^2$

= $\frac{8×7}{2×1}(64x^{12})(\frac{9}{4x^2})$

= 4032 x¹⁰

问题 9. 求系数：

(i) x ¹⁰在 (2x ² – 1/x) ²⁰的扩展中

 解决方案：      

We are given, (2x² – 1/x)²⁰

We know, the (r+1)^th term of the expression is given by,

T_r+1 = ⁿC_r x^n-r a^r

= ²⁰C_r (2x²)^20-r (-1/x)^r

= (-1)^{r 20}C_r (2)^20-r x^40-2r-r

If x¹⁰ exists in the expansion, we must have,

=> 40 − 3r = 10

=> 3r = 30

=> r = 10

Coefficient of x¹⁰ = (-1)^{10 20}C₁₀ (2)^20-10

= ²⁰C₁₀ (2)¹⁰

(ii) x ⁷在 (x – 1/x ² ) ⁴⁰的扩展中

解决方案：

We are given, (x – 1/x²)⁴⁰

We know, the (r+1)^th term of the expression is given by,

T_r+1 = ⁿC_r x^n-r a^r

= ⁴⁰C_r (x)^40-r (-1/x²)^r

= (-1)^{r 40}C_r x^40-r-2r

If x⁷ exists in the expansion, we must have,

=> 40 − 3r = 7

=> 3r = 33

=> r = 11

Coefficient of x⁷ = (-1)^{11 40}C₁₁ 

= − ⁴⁰C₁₁ 

(iii) x ^-15在 (3x ² – a/3x ³ ) ¹⁰的扩展中

解决方案：

We are given, (3x² – a/3x³)¹⁰

We know, the (r+1)^th term of the expression is given by,

T_r+1 = ⁿC_r x^n-r a^r

= ¹⁰C_r (3x²)^10-r (a/3x³)^r

= (-1)^{r 10}C_r 3^10-r-r x^20-2r-3r a^r

If x^-15 exists in the expansion, we must have,

=> 20 − 5r = −15

=> 5r = 35

=> r = 7

Coefficient of x^-15 = (-1)^{7 10}C₇ 3^10-14 a⁷

= $-\frac{10×9×8}{3×2×9×9}a^7$

= −40a⁷/27

(iv) x ⁹在 (x ² – 1/3x) ⁹的扩展中

解决方案：

We are given, (x² – 1/3x)⁹

We know, the (r+1)^th term of the expression is given by,

T_r+1 = ⁿC_r x^n-r a^r

= ⁹C_r (x²)^9-r (-1/3x)^r

= (-1)^{r 9}C_r x^18-2r-r (1/3)^r

If x⁹ exists in the expansion, we must have,

=> 18 − 3r = 9

=> 3r = 9

=> r = 3

Coefficient of x⁹ = (-1)^{3 9}C₃ (1/3)³

= $-\frac{9×8×7}{2×9×9}$

= −28/9

(v) x ^m在 (x + 1/x) ⁿ的展开式中

解决方案：

We are given, (x + 1/x)ⁿ

We know, the (r + 1)^th term of the expression is given by,

T_r+1 = ⁿC_r x^{n- r} (1/x^r)

= ⁿC_r x^{n- 2r} 

If x^m exists in the expansion, we must have,

=> n – 2r = m

=> r = (n – m)/2

Coefficient of x^m = ⁿC_(n-m)/2

= $\frac{n!}{\left( \frac{n - m}{2} \right)! \left( \frac{n + m}{2} \right)!}$

(vi) x 在 (1 – 2x ³ + 3x ⁵ ) (1 + 1/x) ⁸的展开式中

解决方案：

We are given, (1 – 2x³ + 3x⁵) (1 + 1/x)⁸

We know, the (r + 1)^th term of the expression is given by,

= (1 – 2x³ + 3x⁵) (⁸C₀ + ⁸C₁ (1/x) + ⁸C₂ (1/x²) + ⁸C₃ (1/x³) + ⁸C₄ (1/x⁴) + ⁸C₅ (1/x⁵) + ⁸C₆ (1/x⁶) + ⁸C₇ (1/x⁷) + ⁸C₈ (1/x⁸))

Here, x appear in the above expression at -2 x^{3 8}C₂ (1/x²) + 3x⁵.⁸C₄ (1/x⁴)

So, coefficient of x = $-2\left( \frac{8!}{2! 6!} \right) + 3\left( \frac{8!}{4! 4!} \right)$

= – 56 + 210 

= 154

(vii) (a – 2b) ¹²的扩展中的 a ⁵ b ⁷

解决方案：

We are given, (a – 2b)¹²

We know, the (r+1)^th term of the expression is given by,

T_r+1 = ⁿC_r x^n-r a^r

= (-1)^{r 12}C_r (a)^12-r (2b)^r

If a⁵b⁷ exists in the expansion, we must have,

=> 12 − r = 5

=> r = 7

Coefficient of a⁵b⁷ = (-1)^{7 12}C₇ (2)⁷

= $-\frac{12×11×10×9×8×128}{5×4×3×2}$

= − 101376

(viii) x 在 (1 – 3x + 7 x ² ) (1 – x) ¹⁶的展开式中

解决方案：

We are given, (1 – 3x + 7 x²) (1 – x)¹⁶

We know, the (r + 1)^th term of the expression is given by,

= (1 – 3x + 7x²) (¹⁶C₀ + ¹⁶C₁ (-x) + ¹⁶C₂ (-x)² + ¹⁶C₃ (-x)³+ ¹⁶C₄ (-x)⁴ + ¹⁶C₅ (-x)⁵ + ¹⁶C₆ (-x)⁶ + ¹⁶C₇ (-x)⁷ + ¹⁶C₈ (-x)⁸ + ¹⁶C₉ (-x)⁹ + ¹⁶C₁₀ (-x)¹⁰ + ¹⁶C₁₁ (-x)¹¹ + ¹⁶C₁₂ (-x)12 + ¹⁶C₁₃ (-x)¹³ + ¹⁶C₁₄ (-x)¹⁴ + ¹⁶C₁₅ (-x)¹⁵ + ¹⁶C₁₆ (-x)¹⁶)

Here, x appear in the above expression at ¹⁶C₁ (-x) – 3x ¹⁶C₀

So, the coefficient of x = $-\left( \frac{16!}{1! 15!} \right) - 3\left( \frac{16!}{0! 16!} \right)$

= – 16 – 3 

= – 19

问题 10. 展开式中的哪个词$\left[\left(\frac{x}{\sqrt{y}}\right)^{\frac{1}{3}}+\left(\frac{x}{\sqrt[3]{y}}\right)^{\frac{1}{2}}\right]^{21}$包含 x 和 y 的 1 次幂。

解决方案：

We are given, $\left[\left(\frac{x}{\sqrt{y}}\right)^{\frac{1}{3}}+\left(\frac{x}{\sqrt[3]{y}}\right)^{\frac{1}{2}}\right]^{21}$

We know, the (r+1)^th term of the expression is given by,

T_r+1 = ⁿC_r x^n-r a^r

= $^{21}C_r\left[(\frac{x}{\sqrt{y}})^{\frac{1}{3}}\right]^{21-r}\left[(\frac{y}{\sqrt[3]x})^{\frac{1}{2}}\right]^{r}$

= ²¹C_r x^7-r/2 y^2r/3-7/2 

If x and y have same power, we must have,

=> 7 − r/2 = 2r/3 − 7/2

=> 7r/6 = 21/2

=> r = 9

Hence the required term is 9 + 1 = 10^th term.

问题 11. (2x ² – 1/x) ²⁰的展开式是否包含任何涉及 x ⁹的项？

解决方案：

We are given, (2x² – 1/x)²⁰

We know, the (r+1)^th term of the expression is given by,

T_r+1 = ⁿC_r x^n-r a^r

= ²⁰C_r (2x²)^20-r (1/x)^r

= ²⁰C_r (2)^20-r x^40-2r-r 

If x⁹ exists in the expansion, we must have,

=> 40 − 3r = 9

=> 3r = 31

=> r = 31/3

It is not possible, since r is not an integer.

Hence, there is no term with x⁹ in the given expansion.

问题 12. 证明 (x ² + 1/x) ¹²的展开式不包含任何涉及 x ^-1的项。

解决方案：

We have, (x² + 1/x)¹²

We know, the (r+1)^th term of the expression is given by,

T_r+1 = ⁿC_r x^n-r a^r

= ¹²C_r (x²)^12-r (1/x)^r

= ¹²C_r x^24-2r-r

For this term to contain x^-1, we must have

=> 24 – 3r = −1

=> 3r = 24 + 1

=> 3r = 25

=> r = 25/3

It is not possible, since r is not an integer.

Hence, there is no term with x^-1 in the given expansion.

问题 13. 找出展开式中的中间项：

(i) (2/3x – 3/2x) ²⁰

解决方案：

We have,

(2/3x – 3/2x)²⁰ where, n = 20 (which is an even number)

So, the middle term is (n/2 + 1) = (20/2 + 1) = (10 + 1) = 11^th term

Now,

T₁₁ = T₁₀₊₁

= ²⁰C₁₀ (2/3x)^20-10 (3/2x)¹⁰

= ²⁰C₁₀ (2¹⁰/3¹⁰) × (310/210) x^10-10

= ²⁰C₁₀

(ii) (x ² – 2/x) ¹⁰

解决方案：

We have,

(x² – 2/x)¹⁰ where, n = 10 (which is an even number)

So, the middle term is (n/2 + 1) = (10/2 + 1) = (5 + 1) = 6^th term

Now,

T₆ = T₅₊₁

= ¹⁰C₅ (x²)^10-5 (-2/x)⁵

= $-\frac{10×9×8×7×6}{5×4×3×2×1}×32x^5$

= − 8064 x⁵

(iii) (x/a – a/x) ¹⁰

解决方案：

We have, 

(x/a – a/x)¹⁰ where, n = 10 (even number).

So the middle term is (n/2 + 1) = (10/2 + 1) = (5 + 1) = 6^th term

Now,

T₆ = T₅₊₁

= ¹⁰C₅ (x/a)^10-5 (-a/x)⁵

= $-\frac{10×9×8×7×6}{5×4×3×2×1}$

= −252