Binomial Theorem Class 11
In binomial theorem class 11, chapter 8 provides the information regarding the introduction and basic definitions for binomial theorem in a detailed way. To score good marks in binomial theorem class 11 concepts, go through the given problems here. Solve all class 11 Maths Chapter 8 problems in the book by referring to the examples to clarify your binomial theorem concepts.
Binomial Theorem Class 11 Topics
The topics and sub-topics covered in binomial theorem class 11 are:
- Introduction
- Binomial theorem for positive integral indices
- Binomial theorem for any positive integer n
- Special Cases
- General and Middle Term
Binomial Theorem Class 11 Notes
The binomial theorem states a formula for the expression of the powers of sums. The most succinct version of this formula is shown immediately below:
[latex](x+y)^r=\sum_{k=0}^{\infty}\binom{r}{k}x^{r-k}y^k[/latex]
From the above representation, we can expand (a + b)^{n} as given below:
(a + b)^{n} = ^{n}C_{0} a^{n} + ^{n}C_{1} a^{n-1} b + ^{n}C_{2} a^{n-2} b^{2} + … + ^{n}C_{n-1} a b^{n-1} + ^{n}C_{n} b^{n}
This is the binomial theorem formula for any positive integer n.
Some special cases from the binomial theorem can be written as:
- (x + y)^{n} = ^{n}C_{0} x^{n} + ^{n}C_{1} x^{n-1} by+ ^{n}C_{2} x^{n-2} y^{2} + … + ^{n}C_{n-1} x y^{n-1} + ^{n}C_{n} x^{n}
- (x – y)^{n} = ^{n}C_{0} x^{n} – ^{n}C_{1} x^{n-1} by + ^{n}C_{2} x^{n-2} y^{2} + … + (-1)^{n} ^{n}C_{n} x^{n}
- (1 – x)^{n} = ^{n}C_{0} – ^{n}C_{1} x + ^{n}C_{2} x^{2} – …. (-1)^{n} ^{n}C_{n} x^{n}
Also, ^{n}C_{0} = ^{n}C_{n} = 1
However, there will be (n + 1) terms in the expansion of (a + b)^{n}.
General and Middle terms
Consider the binomial expansion, (a + b)^{n} = ^{n}C_{0} a^{n} + ^{n}C_{1} a^{n-1} b + ^{n}C_{2} a^{n-2} b^{2} + … + ^{n}C_{n-1} a b^{n-1} + ^{n}C_{n} b^{n}
Here,
First term = ^{n}C_{0} a^{n}
Second term = ^{n}C_{1} a^{n-1} b
Third term = ^{n}C_{2} a^{n-2} b^{2}
Similarly, we can write the (r + 1)th term as:
^{n}C_{r} a^{n-r} b^{r}
This is the general term of the given expansion.
Thus, (r + 1)Th term, i.e. T_{r+1} = ^{n}C_{r} a^{n-r} b^{r} is called the middle term of the expansion (a + b)^{n}.
Learn more about the general and middle terms of the binomial expansion here.
Binomial Theorem Class 11 Examples
Example 1: Expand: [x^{2} + (3/x)]^{4}, x ≠ 0
Solution:
[x^{2} + (3/x)]^{4}
Using binomial theorem,
[x^{2} + (3/x)]^{4} = ^{4}C_{0} (x^{2})^{4} + ^{4}C_{1} (x^{2})^{3} (3/x) + ^{4}C_{2} (x^{2})^{2} (3/x)^{2} + ^{4}C_{3} (x^{2}) (3/x)^{3} + ^{4}C_{4} (3/x)^{4}
= x^{8} + 4 x^{6} (3/x) + 6 x^{4} (9/x^{2}) + 4 x^{2} (27/x^{3}) + (81/x^{4})
= x^{8} + 12x^{5} + 54x^{2} + (108/x) + (81/x^{4})
Example 2: Compute (98)^{5}
Solution:
Let us write the number 98 as the difference between the two numbers.
98 = 100 – 2
So, (98)^{5} = (100 – 2)^{5}
Using binomial expansion,
(98)^{5} = ^{5}C_{0} (100)^{5} – ^{5}C_{1} (100)^{4} (2) + ^{5}C_{2} (100)^{3} (2)^{2} – ^{5}C_{3} (100)^{2} (2)^{3} + ^{5}C_{4} (100) (2)^{4} – ^{5}C_{5} (2)^{5}
= 10000000000 – 5 × 100000000 × 2 + 10 × 1000000 × 4 – 10 ×10000 × 8 + 5 × 100 × 16 – 32
= 10040008000 – 1000800032
= 9039207968
Example 3: Find the coefficient of x^{6}y^{3} in the expansion (x + 2y)^{9}.
Solution:
Let x^{6}y^{3} be the (r + 1)th term of the expansion (x + 2y)^{9}.
So,
T_{r+1} = ^{9}C_{r} x^{9-r} (2y)^{r}
x^{6}y^{3} = ^{9}C_{r} x^{9-r} 2^{r} y^{r}
By comparing the indices of x and y, we get r = 3.
Coefficient of x^{6}y^{3} = ^{9}C_{3} (2)^{3}
= 84 × 8
= 672
Therefore, the coefficient of x^{6}y^{3} in the expansion (x + 2y)^{9} is 672.
Example 4: The second, third and fourth terms in the binomial expansion (x + a)^{n} are 240, 720 and 1080, respectively. Find x, a and n.
Solution:
Given,
Second term = T_{2} = 240
Third term = T_{3} = 720
Fourth term = T_{4} = 1080
Now,
T_{2} = T_{1+1} = ^{n}C_{1} x^{n-1} (a)
^{n}C_{1} x^{n-1} a = 240….(i)
Similarly,
^{n}C_{2} x^{n-2} a^{2} = 720….(ii)
^{n}C_{3} x^{n-3} a^{3} = 1080….(iii)
Dividing (ii) by (i),
[^{n}C_{2} x^{n-2} a^{2}]/ [^{n}C_{1} x^{n-1} a] = 720/240
[(n – 1)!/(n – 2)!].(a/x) = 6
(n – 1) (a/x) = 6
a/x = 6/(n – 1)….(iv)
Similarly, by dividing (iii) by (ii),
a/x = 9/[2(n – 2)]….(v)
From (iv) and (v),
6/(n – 1) = 9/[2(n – 2)]
12(n – 2) = 9(n – 1)
12n – 24 = 9n – 9
12n – 9n = 24 – 9
3n = 15
n = 5
Subsituting n = 5 in (i),
^{5}C_{1} x^{4} a = 240
ax^{4} = 240/5
ax^{4} = 48….(vi)
Substituting n = 5 in (iv),
a/x = 6/(5 – 1)
a/x = 6/4 = 3/2
a = (3x/2)
Putting this oin equ (vi), we get;
(3x/2) x^{4} = 48
x^{5} = 32
x^{5} = 25
⇒ x = 2
Substituting x = 2 in a = (3x/2)
a = 3(2)/2 = 3
Therefore, x = 2, a = 3 and n = 5.
Practice Problems
- The coefficients of three consecutive terms in the expansion of (1 + a)^{n} are in the ratio1: 7 : 42. Find n.
- What is the middle term in the expansion of [3x – (x^{3}/6)]^{7}?
- If the coefficients of a^{r-1}, a^{r} and a^{r+1} in the expansion of (1 + a)^{n} are in arithmetic progression, prove that n^{2} – n(4r + 1) + 4r^{2} – 2 = 0.
Stay tuned with BYJU’S – The Learning App to gain knowledge on binomial theorem concepts and also refer to more practice problems to score more marks.
Related Links | |
Complex Number Class 11 | Limits and Derivatives Class 11 |
Linear Inequalities Class 11 | Permutation and combination class 11 |