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 = nC0 an + nC1 an-1 b + nC2 an-2 b2 + … + nCn-1 a bn-1 + nCn bn
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 = nC0 xn + nC1 xn-1 by+ nC2 xn-2 y2 + … + nCn-1 x yn-1 + nCn xn
- (x – y)n = nC0 xn – nC1 xn-1 by + nC2 xn-2 y2 + … + (-1)n nCn xn
- (1 – x)n = nC0 – nC1 x + nC2 x2 – …. (-1)n nCn xn
Also, nC0 = nCn = 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 = nC0 an + nC1 an-1 b + nC2 an-2 b2 + … + nCn-1 a bn-1 + nCn bn
Here,
First term = nC0 an
Second term = nC1 an-1 b
Third term = nC2 an-2 b2
Similarly, we can write the (r + 1)th term as:
nCr an-r br
This is the general term of the given expansion.
Thus, (r + 1)Th term, i.e. Tr+1 = nCr an-r br 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: [x2 + (3/x)]4, x ≠ 0
Solution:
[x2 + (3/x)]4
Using binomial theorem,
[x2 + (3/x)]4 = 4C0 (x2)4 + 4C1 (x2)3 (3/x) + 4C2 (x2)2 (3/x)2 + 4C3 (x2) (3/x)3 + 4C4 (3/x)4
= x8 + 4 x6 (3/x) + 6 x4 (9/x2) + 4 x2 (27/x3) + (81/x4)
= x8 + 12x5 + 54x2 + (108/x) + (81/x4)
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 = 5C0 (100)5 – 5C1 (100)4 (2) + 5C2 (100)3 (2)2 – 5C3 (100)2 (2)3 + 5C4 (100) (2)4 – 5C5 (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 x6y3 in the expansion (x + 2y)9.
Solution:
Let x6y3 be the (r + 1)th term of the expansion (x + 2y)9.
So,
Tr+1 = 9Cr x9-r (2y)r
x6y3 = 9Cr x9-r 2r yr
By comparing the indices of x and y, we get r = 3.
Coefficient of x6y3 = 9C3 (2)3
= 84 × 8
= 672
Therefore, the coefficient of x6y3 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 = T2 = 240
Third term = T3 = 720
Fourth term = T4 = 1080
Now,
T2 = T1+1 = nC1 xn-1 (a)
nC1 xn-1 a = 240….(i)
Similarly,
nC2 xn-2 a2 = 720….(ii)
nC3 xn-3 a3 = 1080….(iii)
Dividing (ii) by (i),
[nC2 xn-2 a2]/ [nC1 xn-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),
5C1 x4 a = 240
ax4 = 240/5
ax4 = 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) x4 = 48
x5 = 32
x5 = 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 – (x3/6)]7?
- If the coefficients of ar-1, ar and ar+1 in the expansion of (1 + a)n are in arithmetic progression, prove that n2 – n(4r + 1) + 4r2 – 2 = 0.
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