JEE Main 2025 April

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Animated Solution for Mathematics - Binomial Theorem: Let $(1 + x + x^{2})^{10} = a_{0} + a_{1} x + a_{2} x^{2} + \dots + a_{20} x^{20}$ . If $(a_{1} + a_{3} + a_{5} + \dots + a_{19}) - 11 a_{2} = 121 k$ , then $k$ is equal to .

Q: Wait, why do we use $P(1) - P(-1)$ instead of adding them?

Great question! When you add $P(1)$ and $P(-1)$, the odd-indexed terms ($a_1, a_3, \dots$) cancel out, leaving you with twice the sum of the even-indexed terms. Since our target expression specifically asks for the sum of odd-indexed coefficients, we subtract $P(-1)$ from $P(1)$ to cancel the even terms and double the odd ones.

Q: How do I know which terms in the binomial expansion contribute to $a_2$?

Look at the powers of $x$ in each term of the expansion $\sum \binom{10}{r}(x+x^2)^r$. For $r=1$, we get $10(x+x^2)$, which gives $10x^2$. For $r=2$, we get $45(x+x^2)^2 = 45(x^2 + 2x^3 + x^4)$, which gives $45x^2$. For $r \ge 3$, the lowest power of $x$ is $x^3$, so those terms cannot contribute to $a_2$. It is all about identifying the 'threshold' power!

Enter Numerical Value:

Visualized Solution

Analyzing the Expansion

Given: $(1 + x + x^{2})^{10} = a_{0} + a_{1} x + a_{2} x^{2} + \dots + a_{20} x^{20}$
Target: Find $k$ where $(a_{1} + a_{3} + \dots + a_{19}) - 11 a_{2} = 121 k$

Strategy for Odd Coefficients

To isolate odd-indexed coefficients like $a_{1}, a_{3}, a_{5}$ , we use a standard trick.
We will evaluate the polynomial at $x = 1$ and $x = - 1$ .
Let $P (x) = (1 + x + x^{2})^{10}$ .

Substituting $x = 1$

Put $x = 1$ in the expansion:
$P (1) = (1 + 1 + 1^{2})^{10} = 3^{10}$
$3^{10} = a_{0} + a_{1} + a_{2} + \dots + a_{20} \dots (1)$

Substituting $x = - 1$

Put $x = - 1$ in the expansion:
$P (- 1) = (1 - 1 + (- 1)^{2})^{10} = 1^{10} = 1$
$1 = a_{0} - a_{1} + a_{2} - a_{3} + \dots + a_{20} \dots (2)$

Isolating the Odd Sum

Subtract Equation (2) from Equation (1):
$P (1) - P (- 1) = 3^{10} - 1$
$(a_{0} + a_{1} + \dots) - (a_{0} - a_{1} + \dots) = 2 (a_{1} + a_{3} + a_{5} + \dots + a_{19})$
$2 (a_{1} + a_{3} + \dots + a_{19}) = 3^{10} - 1$

Calculating the Odd Sum

We know $3^{5} = 243$ , so $3^{10} = 243 \times 243 = 59049$ .
$2 (a_{1} + a_{3} + \dots + a_{19}) = 59049 - 1 = 59048$
$a_{1} + a_{3} + \dots + a_{19} = \frac{59048}{2} = 29524$

Strategy for Finding $a_{2}$

We need the coefficient of $x^{2}$ , which is $a_{2}$ .
Treat $(1 + x + x^{2})^{10}$ as a binomial: $(1 + (x + x^{2}))^{10}$ .
Expand using the Binomial Theorem: $\sum_{r = 0}^{10} (r 10) (x + x^{2})^{r}$

Extracting $x^{2}$ Terms

We only need terms up to power 2.
For $r = 0$ : $(0 10) = 1$ (No $x^{2}$ )
For $r = 1$ : $(1 10) (x + x^{2}) \to$ gives $10 x^{2}$
For $r = 2$ : $(2 10) (x + x^{2})^{2} = 45 (x^{2} + 2 x^{3} + x^{4}) \to$ gives $45 x^{2}$
For $r \geq 3$ : Minimum power of $x$ is $x^{3}$ .

Calculating $a_{2}$

Add the coefficients of $x^{2}$ from $r = 1$ and $r = 2$ :
$a_{2} = 10 + 45$
$a_{2} = 55$

Setting Up the Final Equation

Substitute the calculated values into the target equation:
$(a_{1} + a_{3} + \dots + a_{19}) - 11 a_{2} = 121 k$
$29524 - 11 (55) = 121 k$

Solving for $k$

Calculate $11 \times 55 = 605$ .
$29524 - 605 = 121 k$
$28919 = 121 k$
$k = \frac{28919}{121}$

Final Calculation and Conclusion

Divide $28919$ by $121$ :
$k = 239$
Key Takeaway: Use $P (1) - P (- 1)$ for odd coefficients and binomial expansion for specific low-power coefficients.

00:00 / 00:00

The Sigma Insight: Properties of Binomial Coefficients

Analyzing the Setup

The polynomial is given by

P (x) = (1 + x + x^{2})^{10}

. It can be expressed as the sum of its coefficients:

P (x) = a_{0} + a_{1} x + a_{2} x^{2} + \dots + a_{20} x^{20}

Our objective is to determine the value of

k

in the relation:

(a_{1} + a_{3} + \dots + a_{19}) - 11 a_{2} = 121 k

The Odd Sum Mystery

To isolate the sum of the odd-indexed coefficients, we utilize the substitution method by evaluating

P (x)

x = 1

and

x = - 1

x = 1

P (1) = (1 + 1 + 1^{2})^{10} = 3^{10} = 59049

This yields the equation:

a_{0} + a_{1} + a_{2} + \dots + a_{20} = 59049

x = - 1

P (- 1) = (1 - 1 + (- 1)^{2})^{10} = 1^{10} = 1

This yields the equation:

a_{0} - a_{1} + a_{2} - a_{3} + \dots + a_{20} = 1

Subtracting the second equation from the first eliminates the even-indexed terms:

P (1) - P (- 1) = 2 (a_{1} + a_{3} + \dots + a_{19}) = 59049 - 1 = 59048

Dividing by

2

, we find the sum of the odd coefficients:

(a_{1} + a_{3} + \dots + a_{19}) = 29524

The Surgical Extraction of $a_{2}$

To find

a_{2}

without full expansion, we rewrite the polynomial as

(1 + (x + x^{2}))^{10}

and apply the Binomial Theorem:

P (x) = r = 0 \sum 10 (r 10) (x + x^{2})^{r}

We only need the coefficient of

x^{2}

: 1. For

r = 1

, the term is

(1 10) (x + x^{2})

, which contributes

10 x^{2}

. 2. For

r = 2

, the term is

(2 10) (x + x^{2})^{2} = 45 (x^{2} + 2 x^{3} + x^{4})

, which contributes

45 x^{2}

Terms with

r > 2

result in powers of

x

higher than

x^{2}

. Thus, the coefficient

a_{2}

is:

a_{2} = 10 + 45 = 55

The Final Convergence

We now substitute our derived values into the target equation:

29524 - 11 (55) = 121 k

Calculating the product:

11 \times 55 = 605

29524 - 605 = 28919

Finally, we solve for

k

28919 = 121 k

k = \frac{28919}{121} = 239

The final value is $k = 239$ .

Common Traps & FAQs

Wait, why do we use

P (1) - P (- 1)

instead of adding them?

Great question! When you add

P (1)

and

P (- 1)

, the odd-indexed terms (

a_{1}, a_{3}, \dots

) cancel out, leaving you with twice the sum of the even-indexed terms. Since our target expression specifically asks for the sum of odd-indexed coefficients, we subtract

P (- 1)

from

P (1)

to cancel the even terms and double the odd ones.

Why can't we just use the multinomial theorem to find

a_{2}

You absolutely could! The multinomial theorem is powerful, but it often leads to a tedious summation over many indices. By grouping the expression as

(1 + (x + x^{2}))^{10}

, we reduce it to a simple binomial expansion, which is much faster and less prone to calculation errors for low-power coefficients like

a_{2}

How do I know which terms in the binomial expansion contribute to

a_{2}

Look at the powers of

x

in each term of the expansion

\sum (r 10) (x + x^{2})^{r}

. For

r = 1

, we get

10 (x + x^{2})

, which gives

10 x^{2}

. For

r = 2

, we get

45 (x + x^{2})^{2} = 45 (x^{2} + 2 x^{3} + x^{4})

, which gives

45 x^{2}

. For

r \geq 3

, the lowest power of

x

x^{3}

, so those terms cannot contribute to

a_{2}

. It is all about identifying the 'threshold' power!

Animated Solution for Mathematics - Binomial Theorem: Let $(1 + x + x^{2})^{10} = a_{0} + a_{1} x + a_{2} x^{2} + \dots + a_{20} x^{20}$ . If $(a_{1} + a_{3} + a_{5} + \dots + a_{19}) - 11 a_{2} = 121 k$ , then $k$ is equal to .

Enter Numerical Value:

Visualized Solution

Analyzing the Expansion

Strategy for Odd Coefficients

Substituting $x = 1$

Substituting $x = - 1$

Isolating the Odd Sum

Calculating the Odd Sum

Strategy for Finding $a_{2}$

Extracting $x^{2}$ Terms

Calculating $a_{2}$

Setting Up the Final Equation

Solving for $k$

Final Calculation and Conclusion

The Sigma Insight: Properties of Binomial Coefficients

Analyzing the Setup

The Odd Sum Mystery

The Surgical Extraction of $a_{2}$

The Final Convergence

Similar Questions

Let $(1 + x + 2 x^{2})^{20} = a_{0} + a_{1} x + a_{2} x^{2} + \dots + a_{40} x^{40}$ , then $a_{1} + a_{3} + a_{5} + \dots + a_{37}$ is equal to

If $C_{r} =^{25} C_{r}$ and $C_{0} + 5 \cdot C_{1} + 9 \cdot C_{2} + ... + (101) \cdot C_{25} = 2^{25} \cdot k$ , then $k$ is equal to

Let $^{n} C_{r}$ denote the binomial coefficient of $x^{r}$ in the expansion of $(1 + x)^{n}$ . If $\sum_{k = 0}^{10} (2^{2} + 3 k)^{10} C_{k} = α \cdot 3^{10} + β \cdot 2^{10}, α, β \in R$ , then $α + β$ is equal to ___

If $C_{r} =^{25} C_{r}$ and $C_{0} + 5 \cdot C_{1} + 9 \cdot C_{2} + \dots + 101 \cdot C_{25} = 2^{25} \cdot k$ then $k$ is equal to__________.

The value of $(^{21} C_{1} -^{10} C_{1}) + (^{21} C_{2} -^{10} C_{2}) + (^{21} C_{3} -^{10} C_{3}) + (^{21} C_{4} -^{10} C_{4}) + \dots + (^{21} C_{10} -^{10} C_{10})$ is:

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 $\sum_{r = 0}^{10} (\frac{1 0 ^{r + 1} - 1}{1 0 ^{r}}) \cdot^{11} C_{r + 1} = \frac{α ^{11} - 1 1 ^{11}}{1 0 ^{10}}$ , then $α$ is equal to :

Let $S_{n} = 1 + q + q^{2} + \dots + q^{n}$ and $T_{n} = 1 + (\frac{q + 1}{2}) + (\frac{q + 1}{2})^{2} + \dots + (\frac{q + 1}{2})^{n}$ where $q$ is a real number and $q \neq = 1$ . If $^{101} C_{1} +^{101} C_{2} \cdot S_{1} + \dots +^{101} C_{101} \cdot S_{100} = α T_{100}$ then $α$ is equal to:

If $1^{2} \cdot (^{15} C_{1}) + 2^{2} \cdot (^{15} C_{2}) + 3^{2} \cdot (^{15} C_{3}) + \dots + 1 5^{2} \cdot (^{15} C_{15}) = 2^{m} \cdot 3^{n} \cdot 5^{k}$ , where $m, n, k \in N$ , then $m + n + k$ is equal to :

$\sum_{i = 1}^{20} \frac{^{20} C _{i - 1}}{^{20} C _{i} + ^{20} C _{i - 1}} = \frac{k}{21}$ , then k equals :

Enter Numerical Value:

Visualized Solution

Analyzing the Expansion

Strategy for Odd Coefficients

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Substituting .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x=−1

Isolating the Odd Sum

Calculating the Odd Sum

Strategy for Finding .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a2​

Extracting .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x2 Terms

Calculating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a2​

Setting Up the Final Equation

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

Final Calculation and Conclusion

The Sigma Insight: Properties of Binomial Coefficients

Analyzing the Setup

The Odd Sum Mystery

The Surgical Extraction of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a2​

The Final Convergence

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let (1+x+2x2)20=a0​+a1​x+a2​x2+⋯+a40​x40, then a1​+a3​+a5​+⋯+a37​ is equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If Cr​=25Cr​ and C0​+5⋅C1​+9⋅C2​+...+(101)⋅C25​=225⋅k, then k is equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let nCr​ denote the binomial coefficient of xr in the expansion of (1+x)n. If ∑k=010​(22+3k)10Ck​=α⋅310+β⋅210,α,β∈R, then α+β is equal to ___

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If Cr​=25Cr​ and C0​+5⋅C1​+9⋅C2​+⋯+101⋅C25​=225⋅k then k is equal to__________.

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The value of (21C1​−10C1​)+(21C2​−10C2​)+(21C3​−10C3​)+(21C4​−10C4​)+⋯+(21C10​−10C10​) is:

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If ∑r=010​(10r10r+1−1​)⋅11Cr+1​=1010α11−1111​, then α is equal to :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If 12⋅(15C1​)+22⋅(15C2​)+32⋅(15C3​)+⋯+152⋅(15C15​)=2m⋅3n⋅5k, where m,n,k∈N, then m+n+k is equal to :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ∑i=120​20Ci​+20Ci−1​20Ci−1​​=21k​, then k equals :

Substituting $x = 1$

Substituting $x = - 1$

Strategy for Finding $a_{2}$

Extracting $x^{2}$ Terms

Calculating $a_{2}$

Solving for $k$

The Surgical Extraction of $a_{2}$

Let $(1 + x + 2 x^{2})^{20} = a_{0} + a_{1} x + a_{2} x^{2} + \dots + a_{40} x^{40}$ , then $a_{1} + a_{3} + a_{5} + \dots + a_{37}$ is equal to

If $C_{r} =^{25} C_{r}$ and $C_{0} + 5 \cdot C_{1} + 9 \cdot C_{2} + ... + (101) \cdot C_{25} = 2^{25} \cdot k$ , then $k$ is equal to

Let $^{n} C_{r}$ denote the binomial coefficient of $x^{r}$ in the expansion of $(1 + x)^{n}$ . If $\sum_{k = 0}^{10} (2^{2} + 3 k)^{10} C_{k} = α \cdot 3^{10} + β \cdot 2^{10}, α, β \in R$ , then $α + β$ is equal to ___

If $C_{r} =^{25} C_{r}$ and $C_{0} + 5 \cdot C_{1} + 9 \cdot C_{2} + \dots + 101 \cdot C_{25} = 2^{25} \cdot k$ then $k$ is equal to__________.

The value of $(^{21} C_{1} -^{10} C_{1}) + (^{21} C_{2} -^{10} C_{2}) + (^{21} C_{3} -^{10} C_{3}) + (^{21} C_{4} -^{10} C_{4}) + \dots + (^{21} C_{10} -^{10} C_{10})$ is:

If $\sum_{r = 0}^{10} (\frac{1 0 ^{r + 1} - 1}{1 0 ^{r}}) \cdot^{11} C_{r + 1} = \frac{α ^{11} - 1 1 ^{11}}{1 0 ^{10}}$ , then $α$ is equal to :

If $1^{2} \cdot (^{15} C_{1}) + 2^{2} \cdot (^{15} C_{2}) + 3^{2} \cdot (^{15} C_{3}) + \dots + 1 5^{2} \cdot (^{15} C_{15}) = 2^{m} \cdot 3^{n} \cdot 5^{k}$ , where $m, n, k \in N$ , then $m + n + k$ is equal to :

$\sum_{i = 1}^{20} \frac{^{20} C _{i - 1}}{^{20} C _{i} + ^{20} C _{i - 1}} = \frac{k}{21}$ , then k equals :