11类NCERT解决方案–第8章二项式定理–练习8.2

问题1.在(x + 3) ^8中找到x ^5的系数

解决方案：

The (r+1)^th term of (x+3)⁸ is given by T_r+1 = ⁸C_r(x)^8-r(3)^r (eq1).

Therefore for x⁵ we need to get 8-r =5 (Because we need to find x⁵. Therefore, power ox must be equal to 5)

So we get r=3.

Now, put r=3 in eq1. We get,

Coefficient of x⁵ = ⁸C₃(x)⁵(3)³ = 8!*3³/(4!*4!) = 1512

Coefficient of x⁵ is 1512.

问题2.在(a-2b) ^12中找到5 b ⁷的系数。

解决方案：

The (r+1)^th term of (a-2b)¹² is given by T_r+1 = ¹²C_r(a)^12-r(-2b)^r × (eq1)

In the question it is given that exponent of b is 7. Therefore, r should be equal to 7.

By putting r=7 in eq1, we get

Coefficient of a⁵b⁷ = ¹²C₇(-2)⁷ = -101376

问题3.^{在(x 2} -y) ⁶的展开式中写一个通用项。

解决方案：

General term of the equation (a+b)ⁿ is given as T_r+1 = ⁿC_r(a)^n-r(b)^r.

In this question a= x² and b=-y. After putting the value of a, b and n, we get the general term as

T_r+1 = ⁶C_r(x²)^(12-r)(-y)⁶ = (-1)^{r 6}C_r(x)^(12-2r)(y)^r.

问题4.^{在(x 2} -yx) ¹²的展开式中写一个通用项。

解决方案：

General term of the equation (a+b)ⁿ is given as T_r+1 = ⁿC_r(a)^n-r(b)^r.

In this question a= x² and b=-yx. After putting the value of a, b and n, we get the general term as

T_r+1 = ¹²C_r(x²)^(12-r)(-yx)^r = (-1)^{r 12}C_r(x)^(24-2r)(y)^r(x)^r = (-1)^{r 12}C_r(x)^(24-r)(y)^r 

问题5.在(x-2y) ¹²的展开式中找到第四^个项

解决方案：

General term in the expansion of (a+b)ⁿ is written as T_r+1 = ⁿC_r(a)^n-r(b)^r

In the question we are given that a = x, b = -2y and n=12.

To get the 4^th term, we need to put r = 3 (Because r+1=4, therefore r=3).

Therefore, T₄ = ¹²C₃(x)^12-3(-2y)³ = −1760x⁹y³

问题6.在(9x-1 /3√x) ¹⁸的展开式中找到第13^个项

解决方案：

General term in the expansion of (a+b)ⁿ is written as T_r+1 = ⁿC_r(a)^n-r(b)^r

In this question a = 9x, b= -1/3√x and n=18.

To get the 13^th term, we need to put r=12 (Because r+1=13, therefore r=12).

Therefore, T₁₃ = ¹⁸C₁₂(9x)^18-12(-1/3√x)¹² = 18564

问题7.查找的^(3-×3/6)7膨胀中间条款。

解决方案：

In the expansion of (a+b)ⁿ, if n is odd, then there are two middle terms, namely, ((n+1)/2)^th and ((n+1)/2+1)^th term.

Therefore, middle terms in the expansion of (3−x³/6)⁷are 4^th term and 5^th term.

T₄ = T₃₊₁ = ⁷C₃(3)^7-3(−x³/6)³ = (-1)³7!3⁴x⁹/4!.3!6³=-105x⁹/8

T₅ = T₄₊₁ = ⁷C₄(3)^7-4(−x³/6)⁴= (-1)⁴7!3³x¹²/4!3!6⁴ = 35x¹²/48

Thus, the middle terms are -105x⁹/8 and 35x¹²/48.

问题8.在(x / 3 + 9y) ^{10的展开中找到中间项}

解决方案：

In the expansion of (a+b)ⁿ, if n is even, then the middle term is (n/2+1)^th term.

Therefore, the middle term is 6^th term.

T₆ = T₅₊₁ = ¹⁰C₅(x/3)⁵(9y)⁵= (10!x⁵9⁵y⁵)/(5!5!3⁵) = 61236x⁵y⁵

Thus, the middle term is 61236x⁵y⁵

问题9.在(1 ^{+α)M + N，}膨胀证明的^米那系数和^正相等。

解决方案：

Let us assume that a^m occurs in the (r+1)^th term, we obtain

T_r+1 = ^m+nC_r(1)^m+n-r(a)^r = ^m+nC_ra^r

Comparing the indices of a in a^m and in T_r+1, we obtain r=m

Therefore, the coefficient of a^m is ^m+nC_m = (m+n)!/m!n! …(1)

Let us assume that aⁿ occurs in the (k+1)^th, we obtain

T_k+1 = ^m+nC_k(1)^m+n-k(a)^k = ^m+nC_ka^k

Comparing the indices of a in aⁿ and T_k+1, we obtain k-n

Therefore, the coefficient of aⁿ is ^m+nC_n = (m+n)!/m!n! ….(2)

Thus, from (1) and (2), it can be obtained that the coefficients of a^m and aⁿ are equal.

问题10 ：在(x + 1) ⁿ的展开中，^第(r-1)，^第r和第(r + 1)^个项的系数之比为1：3：5。找出n和r。

解决方案：

(r-1)^th term in the expansion of (x+1)ⁿ is T_r-1 = ⁿC_r-2(x)^n-(r-2)(1)^r-2 = ⁿC_r-2(x)^n-r+2

r^th term in the expansion of (x+1)ⁿ is T_r = ⁿC_r-1(x)^n-(r-1)(1)^r-1 = ⁿC_r-1(x)^n-r+1

(r+1)^th term in the expansion of (x+1)ⁿ is T_r+1= ⁿC_r(x)^n-r(1)^r = ⁿC_r(x)^n-r

Therefore the coefficients of (r-1)^th , r^th and (r+1)^th in the expansion of (x+1)ⁿ are ⁿC_r-2, ⁿC_r-1 and ⁿC_r respectively.

Since these coefficients are in the ratio of 1:3:5, we obtain

ⁿC_r-2/ⁿC_r-1 = 1/3 and ⁿC_r-1/ⁿC_r = 3/5

Solving these two equations we get n-4r+5=0 and 3n-8r+3=0.

After solving these two equations, we get n=7 and r=3

Thus n=7 and r=3.

问题11.证明(1 + x) ²ⁿ展开时x ^ n的系数是(1 + x) ^{2n-1展开时x n}的系数的两倍。

解决方案：

In the expansion of (1+x)²ⁿ, T_n+1 = ²ⁿC_n(1)^2n-n(x)ⁿ = ²ⁿC_nxⁿ

Therefore, coefficient of xⁿ in the expansion of (1+x)²ⁿ is ²ⁿC_n

²ⁿC_n = (2n)!/(n!)² ….(1)

Similarly, coefficient of xⁿ in the expansion of (1+x)^2n-1 is ^2n-1C_n

^2n-1C_n = (2n)!/2.(n!)² …(2)

From (1) and (2), 2.^2n-1C_n = ²ⁿC_n

Hence, it is proved that the coefficient of xⁿ in the expansion of (1+x)²ⁿ is twice the coefficient of xⁿ in the expansion of (1+x)^2n-1

问题12。找到一个正值m，在(1 + x) ^{m的展开中，x 2}的系数为6。

解决方案：

General term T_r+1 = ^mC_r(1)^m-r(x)^r = ^mC_r(x)^r

Comparing the indices of x in x² and T_r+1, we get r=2

Therefore, ^mC₂ = 6

$\frac{m!}{(2!(m-2)!)}$= 6

m!/(m-2)!=12

m(m-1) = 12=> m²-m-12 = 0

(m-4)(m+3) = 0

m cannot be negative. Therefore, m=4