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Animated Solution for Mathematics - Binomial Theorem: If the coefficients of $r$ th, $(r + 1)$ th, and $(r + 2)$ th terms in the binomial expansion of $(1 + y)^{m}$ are in A.P., then $m$ and $r$ satisfy the equation

Q: Why do we divide by $m!$ at the start? Is it always safe to do so?

Great question! We divide by $m!$ because it is a common factor across all three terms in our A.P. equation. Since $m$ is a positive integer (the power of the binomial), $m!$ is never zero, so division is perfectly valid. It is a strategic move to reduce the 'noise' of the equation, allowing us to focus on the variables $r$ and $m$ that actually matter.

Select Answer:

Progressive Hints

Hint 1

Hint 1: The Spark: In the binomial expansion of

(1 + y)^{m}

, the general term is

T_{k + 1} =^{m} C_{k} y^{k}

. The most common trap is the index shift. If you are looking for the

r

th term, the index

k

is not

r

, but

r - 1

. Visualize the sequence:

T_{r}

T_{r + 1}

, and

T_{r + 2}

. What are the corresponding values of

k

for each?

Hint 2

Hint 2: The Arsenal: You are given that the coefficients are in Arithmetic Progression (A.P.). Recall the fundamental property of any three numbers

a, b, c

in A.P.: the middle term is the arithmetic mean of the other two, which translates to

2 b = a + c

. Apply this directly to your coefficients:

^{m} C_{r - 1}

^{m} C_{r}

, and

^{m} C_{r + 1}

Hint 3

Hint 3: The Execution: Write out the combination formula

^{n} C_{k} = \frac{n !}{k ! ( n - k )!}

for each term. You will notice

m!

appears in every numerator. Divide the entire equation by

m!

to simplify. Then, to clear the denominators, multiply the entire equation by the product of the largest factorials present:

(r + 1)! (m - r + 1)!

. This will collapse the complex fractions into a clean polynomial equation.

Visualized Solution

Understanding the Binomial Expansion

Given binomial expression: $(1 + y)^{m}$
General term formula: $T_{k + 1} =^{m} C_{k} y^{k}$
We need to find the coefficients of the $r$ th, $(r + 1)$ th, and $(r + 2)$ th terms.

Identifying the Coefficients

For the $r$ th term ( $T_{r}$ ), set $k = r - 1$ : Coefficient is $^{m} C_{r - 1}$
For the $(r + 1)$ th term ( $T_{r + 1}$ ), set $k = r$ : Coefficient is $^{m} C_{r}$
For the $(r + 2)$ th term ( $T_{r + 2}$ ), set $k = r + 1$ : Coefficient is $^{m} C_{r + 1}$

Applying the Arithmetic Progression Condition

Since the coefficients are in A.P.:
Condition: $2 \times (Middle Term) = First Term + Third Term$
Equation: $2 \times^{m} C_{r} =^{m} C_{r - 1} +^{m} C_{r + 1}$

Converting Combinations to Factorials

Using the formula: $^{n} C_{k} = \frac{n !}{k ! ( n - k )!}$
Expanding the equation:
$2 \times \frac{m !}{r ! ( m - r )!} = \frac{m !}{( r - 1 )! ( m - r + 1 )!} + \frac{m !}{( r + 1 )! ( m - r - 1 )!}$

Simplifying the Numerator

Divide both sides of the equation by $m!$ :
$\frac{2}{r ! ( m - r )!} = \frac{1}{( r - 1 )! ( m - r + 1 )!} + \frac{1}{( r + 1 )! ( m - r - 1 )!}$

Analyzing the Factorial Relations

Express larger factorials in terms of smaller ones:
$(r + 1)! = (r + 1) \times r \times (r - 1)!$
$(m - r + 1)! = (m - r + 1) \times (m - r) \times (m - r - 1)!$
We will multiply the entire equation by $(r + 1)! (m - r + 1)!$ to clear denominators.

Clearing the Denominators

After multiplying by $(r + 1)! (m - r + 1)!$ :
Left Side: $2 (r + 1) (m - r + 1)$
First Term on Right: $(r + 1) r$
Second Term on Right: $(m - r + 1) (m - r)$

Expanding the Brackets

Left Side: $2 (m r - r^{2} + r + m - r + 1) = 2 (m r - r^{2} + m + 1)$
Right Side: $(r^{2} + r) + (m^{2} - m r - m r + r^{2} + m - r)$
Simplifying both sides:
$2 m r - 2 r^{2} + 2 m + 2 = 2 r^{2} + m^{2} - 2 m r + m$

Grouping Terms to Match the Options

Move all terms to one side:
$m^{2} - 2 m r - 2 m r - 2 m + m + 2 r^{2} + 2 r^{2} - 2 = 0$
Combine like terms:
$m^{2} - m (4 r + 1) + 4 r^{2} - 2 = 0$

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The Sigma Insight: Properties of Binomial Coefficients

Analyzing the Setup

In the expansion of

(1 + y)^{m}

, the general term is defined as

T_{k + 1} =^{m} C_{k} y^{k}

. Note the index shift: the term number is

k + 1

, while the coefficient index is

k

Consequently, the

r

th term has the coefficient

^{m} C_{r - 1}

, the

(r + 1)

th term has

^{m} C_{r}

, and the

(r + 2)

th term has

^{m} C_{r + 1}

. Mastering this mapping is essential to prevent algebraic errors.

The Master Equation

We are given that these coefficients are in Arithmetic Progression. This implies that the middle term is the arithmetic mean of its neighbors:

2 \times^{m} C_{r} =^{m} C_{r - 1} +^{m} C_{r + 1}

To solve this, we expand each combination using the definition

^{n} C_{k} = \frac{n !}{k ! ( n - k )!}

. Since

m!

appears in every numerator, we divide the entire equation by

m!

to obtain:

\frac{2}{r ! ( m - r )!} = \frac{1}{( r - 1 )! ( m - r + 1 )!} + \frac{1}{( r + 1 )! ( m - r - 1 )!}

Simplifying the Expression

To clear the denominators, we multiply the entire equation by

(r + 1)! (m - r + 1)!

. This surgical simplification yields the following linear-looking equation:

2 (r + 1) (m - r + 1) = (r + 1) r + (m - r + 1) (m - r)

Expanding the left side gives

2 (m r - r^{2} + r + m - r + 1)

, which simplifies to

2 (m r - r^{2} + m + 1)

. Expanding the right side results in

r^{2} + r + m^{2} - 2 m r + r^{2} + m - r

Final Calculation

By carefully combining all terms and moving them to one side of the equation, we arrive at the final quadratic form:

m^{2} - m (4 r + 1) + 4 r^{2} - 2 = 0

This quadratic equation in $m$ represents the relationship between the index

r

and the power

m

for the binomial coefficients to exist in an Arithmetic Progression. You have successfully navigated the complexity to reveal this underlying structure.

Common Traps & FAQs

Wait, why is the

r

th term coefficient

^{m} C_{r - 1}

? Shouldn't it be

^{m} C_{r}

This is the classic 'off-by-one' error that catches even the best students! Remember, the expansion starts with

T_{1} =^{m} C_{0} y^{0}

. Since the index of the combination is always one less than the term number, for the

r

th term, the index must be

r - 1

. Always write out the first two terms to verify your index before proceeding.

Can we use the property

^{n} C_{r} +^{n} C_{r + 1} =^{n + 1} C_{r + 1}

to solve this faster?

That is a brilliant observation! Yes, you absolutely can. Using the Pascal's Identity property can often bypass some of the heavy algebraic lifting. However, for this specific problem, the standard factorial expansion is often safer to ensure you don't miss any terms when rearranging into the final quadratic form. Both paths lead to the same beautiful destination!

Why do we divide by

m!

at the start? Is it always safe to do so?

Great question! We divide by

m!

because it is a common factor across all three terms in our A.P. equation. Since

m

is a positive integer (the power of the binomial),

m!

is never zero, so division is perfectly valid. It is a strategic move to reduce the 'noise' of the equation, allowing us to focus on the variables

r

and

m

that actually matter.

Animated Solution for Mathematics - Binomial Theorem: If the coefficients of $r$ th, $(r + 1)$ th, and $(r + 2)$ th terms in the binomial expansion of $(1 + y)^{m}$ are in A.P., then $m$ and $r$ satisfy the equation

Select Answer:

Visualized Solution

Understanding the Binomial Expansion

Identifying the Coefficients

Applying the Arithmetic Progression Condition

Converting Combinations to Factorials

Simplifying the Numerator

Analyzing the Factorial Relations

Clearing the Denominators

Expanding the Brackets

Grouping Terms to Match the Options

The Sigma Insight: Properties of Binomial Coefficients

Analyzing the Setup

The Master Equation

Simplifying the Expression

Final Calculation

Similar Questions

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For $r = 0, 1, ..., 10$ , let $A_{r}, B_{r}$ and $C_{r}$ denote, respectively, the coefficient of $x^{r}$ in the expansions of $(1 + x)^{10}, (1 + x)^{20}$ and $(1 + x)^{30}$ . Then $\sum_{r = 1}^{10} A_{r} (B_{10} B_{r} - C_{10} A_{r})$ is equal to

If $C_{r}$ stands for $^{n} C_{r}$ , then the sum of the series $\frac{2 ( \frac{n}{2} )! ( \frac{n}{2} )!}{n !} [C_{0}^{2} - 2 C_{1}^{2} + 3 C_{2}^{2} - .... + (- 1)^{n} (n + 1) C_{n}^{2}]$ , where $n$ is an even positive integer, is equal to

Select Answer:

Visualized Solution

Understanding the Binomial Expansion

Identifying the Coefficients

Applying the Arithmetic Progression Condition

Converting Combinations to Factorials

Simplifying the Numerator

Analyzing the Factorial Relations

Clearing the Denominators

Expanding the Brackets

Grouping Terms to Match the Options

The Sigma Insight: Properties of Binomial Coefficients

Analyzing the Setup

The Master Equation

Simplifying the Expression

Final Calculation

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