Welcome, aspiring engineers and mathematicians! Today, we're going to take a deep dive into one of the most fundamental and powerful topics in algebra: the
Binomial Theorem for a Positive Integral Index. This theorem isn't just a formula; it's a doorway to understanding combinations, probability, and many advanced mathematical concepts. Let's build a strong foundation, starting from the very basics.
### The Need for the Binomial Theorem
You're all familiar with expanding expressions like $(a+b)^2 = a^2 + 2ab + b^2$ and $(a+b)^3 = a^3 + 3a^2b + 3ab^2 + b^3$. These are simple enough. But what if I asked you to expand $(a+b)^{10}$ or $(a+b)^{100}$? Manually multiplying this out would be a tedious, error-prone, and time-consuming task. This is where the Binomial Theorem comes to our rescue! It provides a systematic way to expand any binomial (an expression with two terms) raised to any positive integer power.
#### A Glimpse at the Pattern
Let's look at the expansions we know and try to spot a pattern:
* $(a+b)^0 = 1$
* $(a+b)^1 = 1a + 1b$
* $(a+b)^2 = 1a^2 + 2ab + 1b^2$
* $(a+b)^3 = 1a^3 + 3a^2b + 3ab^2 + 1b^3$
* $(a+b)^4 = 1a^4 + 4a^3b + 6a^2b^2 + 4ab^3 + 1b^4$ (You might recall this from school or by multiplying $(a+b)^3$ by $(a+b)$)
Observe the following:
1.
Number of terms: For $(a+b)^n$, there are $(n+1)$ terms.
2.
Powers of 'a' and 'b': The power of 'a' decreases from 'n' to 0, while the power of 'b' increases from 0 to 'n'.
3.
Sum of powers: In every term, the sum of the powers of 'a' and 'b' is always 'n'. For example, in $(a+b)^3$, terms are $a^3b^0, a^2b^1, a^1b^2, a^0b^3$. The sum of powers is $3+0=3$, $2+1=3$, $1+2=3$, $0+3=3$.
4.
Coefficients: The coefficients follow a distinct pattern, which many of you might recognize as
Pascal's Triangle.
### Pascal's Triangle: The Visual Aid
Pascal's Triangle is a geometric arrangement of the binomial coefficients. Each number is the sum of the two numbers directly above it.
1 (n=0)
1 1 (n=1)
1 2 1 (n=2)
1 3 3 1 (n=3)
1 4 6 4 1 (n=4)
1 5 10 10 5 1 (n=5)
This triangle provides the coefficients for $(a+b)^n$ for small values of 'n'. For example, for $(a+b)^4$, the coefficients are 1, 4, 6, 4, 1.
While Pascal's Triangle is excellent for understanding the pattern and for small 'n', it becomes impractical for large values of 'n'. Imagine building it for $n=100$! We need a direct method to find these coefficients. This direct method comes from
Combinatorics.
### Deriving the Binomial Theorem (The Combinatorial Approach)
Let's consider the expansion of $(a+b)^n$. We can write this as:
$(a+b)^n = (a+b)(a+b)(a+b) cdots (a+b)$ (n times)
When we multiply these 'n' binomials, we pick either 'a' or 'b' from each bracket and multiply them.
For example, to get a term like $a^n$, we must choose 'a' from all 'n' brackets. There's only one way to do this: $inom{n}{n}$ or $inom{n}{0}$ (which is 1).
To get a term like $a^{n-1}b^1$, we must choose 'b' from one bracket and 'a' from the remaining $(n-1)$ brackets. How many ways can we choose one 'b' out of 'n' brackets? This is $inom{n}{1}$.
Similarly, to get a term like $a^{n-2}b^2$, we choose 'b' from two brackets and 'a' from the remaining $(n-2)$ brackets. The number of ways to choose two 'b's out of 'n' brackets is $inom{n}{2}$.
In general, to obtain a term $a^{n-k}b^k$, we need to select 'b' from 'k' of the 'n' brackets, and 'a' from the remaining $(n-k)$ brackets. The number of ways to do this is given by the combination formula:
$inom{n}{k} = frac{n!}{k!(n-k)!}$
So, the coefficient of $a^{n-k}b^k$ in the expansion of $(a+b)^n$ is $inom{n}{k}$.
### The Binomial Theorem for a Positive Integral Index
Putting it all together, the Binomial Theorem states:
For any positive integer $n$, the expansion of $(a+b)^n$ is given by:
$mathbf{(a+b)^n = inom{n}{0}a^n b^0 + inom{n}{1}a^{n-1}b^1 + inom{n}{2}a^{n-2}b^2 + cdots + inom{n}{k}a^{n-k}b^k + cdots + inom{n}{n-1}a^1 b^{n-1} + inom{n}{n}a^0 b^n}$
This can be written more compactly using summation notation as:
$mathbf{(a+b)^n = sum_{k=0}^{n} inom{n}{k} a^{n-k} b^k}$
Here:
* $n$ is a positive integer (the index or power).
* $a$ and $b$ are any real or complex numbers.
* $inom{n}{k}$ (read as "n choose k" or $^nC_k$) are the
binomial coefficients, calculated as $frac{n!}{k!(n-k)!}$.
* $k$ is the index of summation, ranging from 0 to $n$. It represents the power of 'b' (or the number of times 'b' is chosen).
JEE Main & Advanced Tip: It's crucial to be absolutely comfortable with the combination notation $inom{n}{k}$ and its properties. Remember $inom{n}{k} = inom{n}{n-k}$.
### Key Observations and Properties
1.
Number of Terms: The expansion of $(a+b)^n$ contains $(n+1)$ terms.
2.
Sum of Indices: In each term, the sum of the exponents of $a$ and $b$ is always $n$.
3.
Symmetry of Coefficients: The binomial coefficients are symmetric, meaning $inom{n}{k} = inom{n}{n-k}$. This implies that coefficients of terms equidistant from the beginning and the end are equal. For example, in $(a+b)^4$, $inom{4}{0}=1$ and $inom{4}{4}=1$; $inom{4}{1}=4$ and $inom{4}{3}=4$.
4.
Special Cases:
* For $(a-b)^n$, replace $b$ with $(-b)$ in the formula:
$(a-b)^n = sum_{k=0}^{n} inom{n}{k} a^{n-k} (-b)^k$
This results in alternating signs:
$inom{n}{0}a^n - inom{n}{1}a^{n-1}b + inom{n}{2}a^{n-2}b^2 - cdots + (-1)^n inom{n}{n}b^n$
* For $(1+x)^n$:
$(1+x)^n = sum_{k=0}^{n} inom{n}{k} (1)^{n-k} x^k = sum_{k=0}^{n} inom{n}{k} x^k = inom{n}{0} + inom{n}{1}x + inom{n}{2}x^2 + cdots + inom{n}{n}x^n$
* For $(1-x)^n$:
$(1-x)^n = sum_{k=0}^{n} inom{n}{k} (1)^{n-k} (-x)^k = sum_{k=0}^{n} (-1)^k inom{n}{k} x^k = inom{n}{0} - inom{n}{1}x + inom{n}{2}x^2 - cdots + (-1)^n inom{n}{n}x^n$
5.
Sum of Binomial Coefficients:
Setting $a=1, b=1$ in $(a+b)^n = sum_{k=0}^{n} inom{n}{k} a^{n-k} b^k$, we get:
$(1+1)^n = sum_{k=0}^{n} inom{n}{k} (1)^{n-k} (1)^k implies 2^n = inom{n}{0} + inom{n}{1} + inom{n}{2} + cdots + inom{n}{n}$
The sum of all binomial coefficients for a given $n$ is $2^n$.
6.
Alternating Sum of Binomial Coefficients:
Setting $a=1, b=-1$ in $(a+b)^n = sum_{k=0}^{n} inom{n}{k} a^{n-k} b^k$, we get:
$(1-1)^n = sum_{k=0}^{n} inom{n}{k} (1)^{n-k} (-1)^k implies 0^n = inom{n}{0} - inom{n}{1} + inom{n}{2} - cdots + (-1)^n inom{n}{n}$
For $n > 0$, the alternating sum of binomial coefficients is 0.
### The General Term (or $T_{r+1}$ Term)
The $(k+1)^{th}$ term in the expansion of $(a+b)^n$ is often denoted as $T_{k+1}$ or $T_{r+1}$.
From our formula, $(a+b)^n = sum_{k=0}^{n} inom{n}{k} a^{n-k} b^k$, we can see that the term corresponding to $k$ is the $(k+1)^{th}$ term.
So, the general term is:
$mathbf{T_{r+1} = inom{n}{r} a^{n-r} b^r}$
Why $T_{r+1}$ and not $T_r$? Because the index $r$ starts from $0$ (for the first term). So, $r=0$ gives the 1st term, $r=1$ gives the 2nd term, and so on. Thus, for the $r^{th}$ term, the index would be $(r-1)$. It's more intuitive to use $r$ as the power of $b$, making it the $(r+1)^{th}$ term.
The general term is immensely useful for:
* Finding a specific term in the expansion.
* Finding the coefficient of a particular power of a variable.
* Finding the term independent of a variable.
#### Example 1: Expanding a Binomial
Expand $(2x+3y)^4$.
Here, $n=4$, $a=2x$, $b=3y$.
Using the formula:
$(2x+3y)^4 = inom{4}{0}(2x)^4(3y)^0 + inom{4}{1}(2x)^3(3y)^1 + inom{4}{2}(2x)^2(3y)^2 + inom{4}{3}(2x)^1(3y)^3 + inom{4}{4}(2x)^0(3y)^4$
Let's calculate the coefficients and terms:
$inom{4}{0} = 1$
$inom{4}{1} = 4$
$inom{4}{2} = frac{4 imes 3}{2 imes 1} = 6$
$inom{4}{3} = 4$
$inom{4}{4} = 1$
So, the expansion is:
$= 1 cdot (16x^4) cdot 1 + 4 cdot (8x^3) cdot (3y) + 6 cdot (4x^2) cdot (9y^2) + 4 cdot (2x) cdot (27y^3) + 1 cdot 1 cdot (81y^4)$
$= mathbf{16x^4 + 96x^3y + 216x^2y^2 + 216xy^3 + 81y^4}$
#### Example 2: Finding a Specific Term
Find the $5^{th}$ term in the expansion of $(x^2 - frac{1}{x})^{10}$.
Here, $n=10$, $a=x^2$, $b=-frac{1}{x}$.
For the $5^{th}$ term, we need $T_{r+1} = T_{4+1}$, so $r=4$.
$T_{r+1} = inom{n}{r} a^{n-r} b^r$
$T_5 = inom{10}{4} (x^2)^{10-4} left(-frac{1}{x}
ight)^4$
$T_5 = inom{10}{4} (x^2)^6 left(-frac{1}{x}
ight)^4$
First, calculate $inom{10}{4} = frac{10 imes 9 imes 8 imes 7}{4 imes 3 imes 2 imes 1} = 10 imes 3 imes 7 = 210$.
$T_5 = 210 cdot x^{12} cdot frac{1}{x^4}$
$T_5 = 210 cdot x^{12-4}$
$mathbf{T_5 = 210x^8}$
#### Example 3: Finding the Coefficient of a Specific Power
Find the coefficient of $x^7$ in the expansion of $(3x^2 - frac{1}{x^3})^8$.
Here, $n=8$, $a=3x^2$, $b=-frac{1}{x^3}$.
The general term $T_{r+1}$ is:
$T_{r+1} = inom{8}{r} (3x^2)^{8-r} left(-frac{1}{x^3}
ight)^r$
$T_{r+1} = inom{8}{r} (3)^{8-r} (x^2)^{8-r} (-1)^r (x^{-3})^r$
$T_{r+1} = inom{8}{r} 3^{8-r} (-1)^r x^{2(8-r)} x^{-3r}$
$T_{r+1} = inom{8}{r} 3^{8-r} (-1)^r x^{16-2r-3r}$
$T_{r+1} = inom{8}{r} 3^{8-r} (-1)^r x^{16-5r}$
We want the coefficient of $x^7$, so we set the exponent of $x$ equal to 7:
$16 - 5r = 7$
$5r = 16 - 7$
$5r = 9$
$r = frac{9}{5}$
Since $r$ must be a non-negative integer (0, 1, 2, ..., n), and we got a fractional value, it means there is no term with $x^7$ in this expansion.
Important: If 'r' is not a non-negative integer, then the required term (or power of x) does not exist in the expansion, and its coefficient is 0.
#### Example 4: Finding the Term Independent of x
Find the term independent of $x$ (constant term) in the expansion of $(frac{3}{2}x^2 - frac{1}{3x})^9$.
Here, $n=9$, $a=frac{3}{2}x^2$, $b=-frac{1}{3x}$.
The general term $T_{r+1}$ is:
$T_{r+1} = inom{9}{r} left(frac{3}{2}x^2
ight)^{9-r} left(-frac{1}{3x}
ight)^r$
$T_{r+1} = inom{9}{r} left(frac{3}{2}
ight)^{9-r} (x^2)^{9-r} (-1)^r left(frac{1}{3}
ight)^r (x^{-1})^r$
$T_{r+1} = inom{9}{r} left(frac{3}{2}
ight)^{9-r} (-1)^r left(frac{1}{3}
ight)^r x^{2(9-r)} x^{-r}$
$T_{r+1} = inom{9}{r} left(frac{3}{2}
ight)^{9-r} left(-frac{1}{3}
ight)^r x^{18-2r-r}$
$T_{r+1} = inom{9}{r} left(frac{3}{2}
ight)^{9-r} left(-frac{1}{3}
ight)^r x^{18-3r}$
For the term independent of $x$, the exponent of $x$ must be 0:
$18 - 3r = 0$
$3r = 18$
$r = 6$
Since $r=6$ is a non-negative integer, a term independent of $x$ exists. Substitute $r=6$ back into the coefficient part of $T_{r+1}$:
Term independent of $x = inom{9}{6} left(frac{3}{2}
ight)^{9-6} left(-frac{1}{3}
ight)^6$
$= inom{9}{3} left(frac{3}{2}
ight)^3 left(-frac{1}{3}
ight)^6$ (since $inom{9}{6} = inom{9}{3}$)
$= frac{9 imes 8 imes 7}{3 imes 2 imes 1} cdot frac{3^3}{2^3} cdot frac{(-1)^6}{3^6}$
$= (3 imes 4 imes 7) cdot frac{27}{8} cdot frac{1}{729}$
$= 84 cdot frac{27}{8} cdot frac{1}{729}$
$= frac{84 imes 27}{8 imes 729} = frac{2268}{5832}$
To simplify, $27 imes 27 = 729$. So $27/729 = 1/27$.
$= 84 cdot frac{1}{8} cdot frac{1}{27}$
$= frac{21}{2} cdot frac{1}{27} = frac{7}{2} cdot frac{1}{9} = frac{7}{18}$
The term independent of $x$ is $mathbf{frac{7}{18}}$.
### Middle Term(s)
The concept of middle terms depends on whether $n$ is even or odd.
1.
If $n$ is even:
There is only one middle term.
Its position is $left(frac{n}{2} + 1
ight)^{th}$ term.
So, $r = frac{n}{2}$.
The middle term is $T_{frac{n}{2}+1} = inom{n}{n/2} a^{n/2} b^{n/2}$.
2.
If $n$ is odd:
There are two middle terms.
Their positions are $left(frac{n+1}{2}
ight)^{th}$ term and $left(frac{n+3}{2}
ight)^{th}$ term.
For the first middle term, $r = frac{n-1}{2}$. This is $T_{frac{n+1}{2}} = inom{n}{(n-1)/2} a^{(n+1)/2} b^{(n-1)/2}$.
For the second middle term, $r = frac{n+1}{2}$. This is $T_{frac{n+3}{2}} = inom{n}{(n+1)/2} a^{(n-1)/2} b^{(n+1)/2}$.
#### Example 5: Finding the Middle Term(s)
a) Find the middle term in the expansion of $(2x + frac{1}{x})^{10}$.
Here, $n=10$ (even).
Position of the middle term $= left(frac{10}{2} + 1
ight)^{th} = (5+1)^{th} = 6^{th}$ term.
So, $r=5$.
$a=2x$, $b=frac{1}{x}$.
$T_6 = inom{10}{5} (2x)^{10-5} left(frac{1}{x}
ight)^5$
$T_6 = inom{10}{5} (2x)^5 left(frac{1}{x}
ight)^5$
$inom{10}{5} = frac{10 imes 9 imes 8 imes 7 imes 6}{5 imes 4 imes 3 imes 2 imes 1} = 2 imes 3 imes 2 imes 7 imes 3 = 252$.
$T_6 = 252 cdot (32x^5) cdot (frac{1}{x^5})$
$T_6 = 252 cdot 32 = mathbf{8064}$.
b) Find the middle terms in the expansion of $(x^3 - frac{2}{x^2})^7$.
Here, $n=7$ (odd).
Positions of the middle terms are $left(frac{7+1}{2}
ight)^{th} = 4^{th}$ term and $left(frac{7+3}{2}
ight)^{th} = 5^{th}$ term.
For $T_4$: $r=3$. $a=x^3$, $b=-frac{2}{x^2}$.
$T_4 = inom{7}{3} (x^3)^{7-3} left(-frac{2}{x^2}
ight)^3$
$T_4 = inom{7}{3} (x^3)^4 left(-frac{2}{x^2}
ight)^3$
$inom{7}{3} = frac{7 imes 6 imes 5}{3 imes 2 imes 1} = 35$.
$T_4 = 35 cdot x^{12} cdot left(frac{-8}{x^6}
ight) = -280 x^{12-6} = mathbf{-280x^6}$.
For $T_5$: $r=4$. $a=x^3$, $b=-frac{2}{x^2}$.
$T_5 = inom{7}{4} (x^3)^{7-4} left(-frac{2}{x^2}
ight)^4$
$T_5 = inom{7}{4} (x^3)^3 left(-frac{2}{x^2}
ight)^4$
$inom{7}{4} = inom{7}{3} = 35$.
$T_5 = 35 cdot x^9 cdot left(frac{16}{x^8}
ight) = 35 imes 16 x^{9-8} = mathbf{560x}$.
### Numerical Greatest Term (NGT) - (JEE Advanced Focus)
This concept focuses on finding the term with the largest numerical value in a binomial expansion, not necessarily the largest coefficient (as coefficients can be negative).
Let $T_{r+1}$ and $T_r$ be two consecutive terms in the expansion of $(a+b)^n$. We are interested in $|T_{r+1}|$ compared to $|T_r|$.
$|T_{r+1}| = left| inom{n}{r} a^{n-r} b^r
ight|$
$|T_r| = left| inom{n}{r-1} a^{n-(r-1)} b^{r-1}
ight|$
Consider the ratio:
$left| frac{T_{r+1}}{T_r}
ight| = left| frac{inom{n}{r} a^{n-r} b^r}{inom{n}{r-1} a^{n-r+1} b^{r-1}}
ight|$
$= left| frac{frac{n!}{r!(n-r)!}}{frac{n!}{(r-1)!(n-r+1)!}} cdot frac{a^{n-r} b^r}{a^{n-r+1} b^{r-1}}
ight|$
$= left| frac{(r-1)!(n-r+1)!}{r!(n-r)!} cdot frac{b}{a}
ight|$
$= left| frac{n-r+1}{r} cdot frac{b}{a}
ight| = left( frac{n-r+1}{r}
ight) left| frac{b}{a}
ight|$
For $|T_{r+1}|$ to be the greatest term, it must be greater than or equal to both its preceding term $|T_r|$ and its succeeding term $|T_{r+2}|$.
We typically find $r$ such that $|T_{r+1}| ge |T_r|$, i.e., $left| frac{T_{r+1}}{T_r}
ight| ge 1$.
$left( frac{n-r+1}{r}
ight) left| frac{b}{a}
ight| ge 1$
$frac{n-r+1}{r} ge frac{1}{|b/a|}$
$frac{n-r+1}{r} ge frac{|a|}{|b|}$
$(n-r+1)|b| ge r|a|$
$n|b| - r|b| + |b| ge r|a|$
$n|b| + |b| ge r|a| + r|b|$
$(n+1)|b| ge r(|a|+|b|)$
$r le frac{(n+1)|b|}{|a|+|b|}$
Let $m = frac{(n+1)|b|}{|a|+|b|}$.
*
If $m$ is an integer: Then $|T_m| = |T_{m+1}|$, and both these terms are the numerically greatest terms.
*
If $m$ is not an integer: Let $lfloor m
floor$ be the greatest integer less than or equal to $m$. Then $r = lfloor m
floor$. The term $T_{lfloor m
floor + 1}$ is the numerically greatest term.
JEE Advanced Alert: NGT problems are common in JEE Advanced. Master this derivation and application. Remember to consider absolute values.
#### Example 6: Finding the Numerically Greatest Term
Find the numerically greatest term in the expansion of $(2+3x)^9$ when $x=frac{3}{2}$.
Here, $n=9$, $a=2$, $b=3x$.
We are interested in the term $T_{r+1}$.
The ratio $left| frac{T_{r+1}}{T_r}
ight| = left( frac{n-r+1}{r}
ight) left| frac{b}{a}
ight|$
Substitute $n=9$, $a=2$, $b=3x = 3(frac{3}{2}) = frac{9}{2}$.
$left| frac{T_{r+1}}{T_r}
ight| = left( frac{9-r+1}{r}
ight) left| frac{9/2}{2}
ight|$
$= left( frac{10-r}{r}
ight) left( frac{9}{4}
ight)$
We want to find $r$ such that $left| frac{T_{r+1}}{T_r}
ight| ge 1$:
$left( frac{10-r}{r}
ight) left( frac{9}{4}
ight) ge 1$
$9(10-r) ge 4r$
$90 - 9r ge 4r$
$90 ge 13r$
$r le frac{90}{13}$
$r le 6.92...$
Since $r$ must be an integer, the largest integer value for $r$ satisfying this condition is $r=6$.
This means that $|T_7| ge |T_6|$.
Also, for $r=7$, $r > 6.92...$, so $|T_8| < |T_7|$.
Thus, the numerically greatest term is $T_{r+1} = T_{6+1} = T_7$.
Now, calculate $T_7$:
$T_7 = inom{9}{6} (2)^{9-6} (3x)^6$
$T_7 = inom{9}{3} (2)^3 (3 cdot frac{3}{2})^6$
$T_7 = frac{9 imes 8 imes 7}{3 imes 2 imes 1} cdot 8 cdot left(frac{9}{2}
ight)^6$
$T_7 = 84 cdot 8 cdot frac{9^6}{2^6}$
$T_7 = 84 cdot 8 cdot frac{9^6}{64}$
$T_7 = 84 cdot frac{1}{8} cdot 9^6$
$T_7 = frac{21}{2} cdot 9^6$
$T_7 = frac{21}{2} cdot 531441 = frac{11160261}{2} = mathbf{5580130.5}$
This comprehensive understanding of the Binomial Theorem for a positive integral index forms the bedrock for advanced topics like the Binomial Theorem for any index, multinomial theorem, and various applications in probability and series. Keep practicing these concepts!