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Animated Solution for Mathematics - Binomial Theorem: The natural number $m$ , for which the coefficient of $x$ in the binomial expansion of $(x^{m} + \frac{1}{x ^{2}})^{22}$ is 1540, is

Q: Wait, why do we get two values for $r$? Does the problem have two answers?

That is a brilliant observation! The symmetry property $\binom{n}{r} = \binom{n}{n-r}$ is a fundamental property of Pascal's Triangle. While both $r=3$ and $r=19$ satisfy the coefficient condition, only one will satisfy the *exponent* condition for a natural number $m$. Always test both candidates; one is often a 'decoy' that fails the constraint check.

Q: Why did we reject $m = 7/19$?

The problem explicitly states that $m$ is a natural number. In mathematics, constraints are not just suggestions; they are the boundaries of our universe. Since $7/19$ is not an integer, it cannot be our solution. Never ignore the domain constraints provided in the question—they are your final filter.

Enter Numerical Value:

Visualized Solution

The Binomial Expression

Given expression: $(x^{m} + \frac{1}{x ^{2}})^{22}$
We need to find the natural number $m$ .
Condition: The coefficient of $x^{1}$ in the expansion is $1540$ .

The General Term Formula

To find a specific term, we use the general term formula.
For $(a + b)^{n}$ , the general term is:
$T_{r + 1} = (r n) a^{n - r} b^{r}$

Substituting the Values

Compare $(x^{m} + x^{- 2})^{22}$ with $(a + b)^{n}$ .
$a = x^{m}$ , $b = x^{- 2}$ , $n = 22$
$T_{r + 1} = (r 22) (x^{m})^{22 - r} (x^{- 2})^{r}$

Simplifying the Powers of $x$

Use exponent rules: $(x^{a})^{b} = x^{ab}$
$(x^{m})^{22 - r} = x^{22 m - m r}$
$(x^{- 2})^{r} = x^{- 2 r}$
Combine bases: $x^{22 m - m r - 2 r}$

Equating the Coefficient

The coefficient part of the term is $(r 22)$ .
We are given that the coefficient is $1540$ .
Therefore, $(r 22) = 1540$ .

Solving for $r$

We need to find $r$ such that $(r 22) = 1540$ .
Let's test small values. For $r = 3$ :
$(3 22) = \frac{22 \times 21 \times 20}{3 \times 2 \times 1} = 1540$
By symmetry $(r n) = (n - r n)$ , $r = 19$ is also a solution.

Equating the Power of $x$

We are looking for the coefficient of $x^{1}$ .
The power of $x$ in our general term is $22 m - m r - 2 r$ .
Therefore, $22 m - m r - 2 r = 1$ .

Testing Case 1 ( $r = 3$ )

Substitute $r = 3$ into $22 m - m r - 2 r = 1$ .
$22 m - 3 m - 2 (3) = 1$
$19 m - 6 = 1 \Rightarrow 19 m = 7$
$m = \frac{7}{19}$ (Not a natural number)

Testing Case 2 ( $r = 19$ )

Substitute $r = 19$ into $22 m - m r - 2 r = 1$ .
$22 m - 19 m - 2 (19) = 1$
$3 m - 38 = 1 \Rightarrow 3 m = 39$
$m = 13$ (This is a natural number!)

Final Conclusion

The natural number $m$ is $13$ .
Key Takeaway: Always check all possible values of $r$ and verify against given constraints (like $m \in N$ ).

00:00 / 00:00

The Sigma Insight: General Term and Middle Term

Solution Diagram

Analyzing the Setup

We are tasked with finding the natural number

m

given the binomial expression

(x^{m} + \frac{1}{x ^{2}})^{22}

, where the coefficient of

x^{1}

1540

Expanding this expression directly would result in

23

terms, which is inefficient. Instead, we utilize the General Term Formula for

(a + b)^{n}

, where the

(r + 1)

-th term is defined as:

T_{r + 1} = (r n) a^{n - r} b^{r}

In this specific problem, we identify

a = x^{m}

b = x^{- 2}

, and

n = 22

. Substituting these values, we obtain:

T_{r + 1} = (r 22) (x^{m})^{22 - r} (x^{- 2})^{r}

The Algebra of Exponents

To isolate the power of

x

, we apply the laws of exponents. Simplifying the expression yields:

T_{r + 1} = (r 22) x^{22 m - m r - 2 r}

The problem provides two critical pieces of information: the coefficient

(r 22) = 1540

and the exponent of

x

must be

1

. We must solve for

r

and

m

simultaneously.

Solving for the Binomial Coefficient

We first address the condition

(r 22) = 1540

. Recalling the symmetry property

(r n) = (n - r n)

, we test values for

r

(3 22) = \frac{22 \times 21 \times 20}{3 \times 2 \times 1} = 22 \times 7 \times 10 = 1540

Since

(3 22) = 1540

, by symmetry,

(19 22)

must also equal

1540

. Thus, our potential candidates for

r

are

3

and

19

The Final Verdict

We now apply the exponent condition

22 m - m r - 2 r = 1

to our candidates.

Case 1: $r = 3$

22 m - 3 m - 2 (3) = 1

19 m - 6 = 1

19 m = 7

m = \frac{7}{19}

Since

m

must be a natural number, we reject this result.

Case 2: $r = 19$

22 m - 19 m - 2 (19) = 1

3 m - 38 = 1

3 m = 39

m = 13

The value

m = 13

is a natural number and satisfies all given constraints. Therefore, the required value is $m = 13$ .

Common Traps & FAQs

Wait, why do we get two values for

r

? Does the problem have two answers?

That is a brilliant observation! The symmetry property

(r n) = (n - r n)

is a fundamental property of Pascal's Triangle. While both

r = 3

and

r = 19

satisfy the coefficient condition, only one will satisfy the *exponent* condition for a natural number

m

. Always test both candidates; one is often a 'decoy' that fails the constraint check.

Why did we reject

m = 7/19

The problem explicitly states that

m

is a natural number. In mathematics, constraints are not just suggestions; they are the boundaries of our universe. Since

7/19

is not an integer, it cannot be our solution. Never ignore the domain constraints provided in the question—they are your final filter.

Can we use the general term formula for negative powers like

x^{- 2}

Absolutely! The binomial theorem is incredibly robust. Whether the term is

x^{m}

x^{- 2}

, the algebra of exponents

x^{a} \cdot x^{b} = x^{a + b}

holds firm. Treat the negative sign as part of the term

b

. The theorem doesn't care if the exponent is positive or negative; it only cares about the structure of the binomial.

Animated Solution for Mathematics - Binomial Theorem: The natural number $m$ , for which the coefficient of $x$ in the binomial expansion of $(x^{m} + \frac{1}{x ^{2}})^{22}$ is 1540, is

Enter Numerical Value:

Visualized Solution

The Binomial Expression

The General Term Formula

Substituting the Values

Simplifying the Powers of $x$

Equating the Coefficient

Solving for $r$

Equating the Power of $x$

Testing Case 1 ( $r = 3$ )

Testing Case 2 ( $r = 19$ )

Final Conclusion

The Sigma Insight: General Term and Middle Term

Analyzing the Setup

The Algebra of Exponents

Solving for the Binomial Coefficient

The Final Verdict

Similar Questions

The smallest natural number $n$ , such that the coefficient of $x$ in the expansion of $(x^{2} + \frac{1}{x ^{3}})^{n}$ is $^{n} C_{23}$ , is :

If the co-efficient of $x^{7}$ and $x^{8}$ in the expansion of $(2 + \frac{x}{3})^{n}$ are equal, then the value of $n$ is equal to ___

If the $101 1^{t h}$ term from the end in the binomial expansion of $(\frac{4 x}{5} - \frac{5}{2 x})^{2022}$ is 1024 times $101 1^{t h}$ term from the beginning, then $32∣ x ∣$ is equal to

Suppose A and B are the coefficients of $3 0^{t h}$ and $1 2^{t h}$ terms respectively in the binomial expansion of $(1 + x)^{2 n - 1}$ . If $2 A = 5 B$ , then n is equal to :

The coefficient of $x^{10}$ in the expansion of $(1 + x)^{2} (1 + x^{2})^{3} (1 + x^{3})^{4}$ is equal to :-

If the sum of the coefficients of all the positive powers of $x$ , in the binomial expansion of $(x^{n} + \frac{2}{x ^{5}})^{7}$ is 939, then the sum of all the possible integral values of $n$ is :

Let the coefficients of $x^{- 1}$ and $x^{- 3}$ in the expansion of $(2 x^{1/5} - \frac{1}{x ^{1/5}})^{15}$ , $x > 0$ , be $m$ and $n$ respectively. If $r$ is a positive integer such that $m n^{2} =^{15} C_{r} \cdot 2^{r}$ , then the value of $r$ is equal to____.

If the fourth term in the expansion of $(x + x^{l o g_{2} x})^{7}$ is 4480, then the value of $x$ where $x \in N$ is equal to :

In the expansion of $(32 + \frac{1}{3 3})^{n}, n \in N$ , if the ratio of $1 5^{t h}$ term from the beginning to the $1 5^{t h}$ term from the end is $\frac{1}{6}$ , then the value of $^{n} C_{3}$ is:

If the coefficient of $x^{10}$ in the binomial expansion of $(\frac{x}{5 ^{1/4}} + \frac{5}{x ^{1/3}})^{60}$ is $5^{k} l$ , where $l, k \in N$ and $l$ is coprime to 5, then $k$ is equal to ____.

Animated Solution for Mathematics - Binomial Theorem: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The natural number m, for which the coefficient of x in the binomial expansion of (xm+x21​)22 is 1540, is

Enter Numerical Value:

Visualized Solution

The Binomial Expression

The General Term Formula

Substituting the Values

Simplifying the Powers of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x

Equating the Coefficient

Solving for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } r

Equating the Power of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x

Testing Case 1 ( .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } r=3)

Testing Case 2 ( .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } r=19)

Final Conclusion

The Sigma Insight: General Term and Middle Term

Analyzing the Setup

The Algebra of Exponents

Solving for the Binomial Coefficient

The Final Verdict

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The smallest natural number n, such that the coefficient of x in the expansion of (x2+x31​)n is nC23​, is :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the co-efficient of x7 and x8 in the expansion of (2+3x​)n are equal, then the value of n is equal to ___

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the 1011th term from the end in the binomial expansion of (54x​−2x5​)2022 is 1024 times 1011th term from the beginning, then 32∣x∣ is equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Suppose A and B are the coefficients of 30th and 12th terms respectively in the binomial expansion of (1+x)2n−1. If 2A=5B, then n is equal to :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The coefficient of x10 in the expansion of (1+x)2(1+x2)3(1+x3)4 is equal to :-

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the sum of the coefficients of all the positive powers of x, in the binomial expansion of (xn+x52​)7 is 939, then the sum of all the possible integral values of n is :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the fourth term in the expansion of (x+xlog2​x)7 is 4480, then the value of x where x∈N is equal to :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } In the expansion of (32​+33​1​)n,n∈N, if the ratio of 15th term from the beginning to the 15th term from the end is 61​, then the value of nC3​ is:

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the coefficient of x10 in the binomial expansion of (51/4x​​+x1/35​​)60 is 5kl, where l,k∈N and l is coprime to 5, then k is equal to ____.

Animated Solution for Mathematics - Binomial Theorem: The natural number $m$ , for which the coefficient of $x$ in the binomial expansion of $(x^{m} + \frac{1}{x ^{2}})^{22}$ is 1540, is

Simplifying the Powers of $x$

Solving for $r$

Equating the Power of $x$

Testing Case 1 ( $r = 3$ )

Testing Case 2 ( $r = 19$ )

The smallest natural number $n$ , such that the coefficient of $x$ in the expansion of $(x^{2} + \frac{1}{x ^{3}})^{n}$ is $^{n} C_{23}$ , is :

If the co-efficient of $x^{7}$ and $x^{8}$ in the expansion of $(2 + \frac{x}{3})^{n}$ are equal, then the value of $n$ is equal to ___

If the $101 1^{t h}$ term from the end in the binomial expansion of $(\frac{4 x}{5} - \frac{5}{2 x})^{2022}$ is 1024 times $101 1^{t h}$ term from the beginning, then $32∣ x ∣$ is equal to

Suppose A and B are the coefficients of $3 0^{t h}$ and $1 2^{t h}$ terms respectively in the binomial expansion of $(1 + x)^{2 n - 1}$ . If $2 A = 5 B$ , then n is equal to :

The coefficient of $x^{10}$ in the expansion of $(1 + x)^{2} (1 + x^{2})^{3} (1 + x^{3})^{4}$ is equal to :-

If the sum of the coefficients of all the positive powers of $x$ , in the binomial expansion of $(x^{n} + \frac{2}{x ^{5}})^{7}$ is 939, then the sum of all the possible integral values of $n$ is :

If the fourth term in the expansion of $(x + x^{l o g_{2} x})^{7}$ is 4480, then the value of $x$ where $x \in N$ is equal to :

In the expansion of $(32 + \frac{1}{3 3})^{n}, n \in N$ , if the ratio of $1 5^{t h}$ term from the beginning to the $1 5^{t h}$ term from the end is $\frac{1}{6}$ , then the value of $^{n} C_{3}$ is:

If the coefficient of $x^{10}$ in the binomial expansion of $(\frac{x}{5 ^{1/4}} + \frac{5}{x ^{1/3}})^{60}$ is $5^{k} l$ , where $l, k \in N$ and $l$ is coprime to 5, then $k$ is equal to ____.