JEE Main 2025 April

LEVELJEE Main

Animated Solution for Mathematics - Binomial Theorem: 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 :

Select Answer:

Visualized Solution

Analyze and Simplify the Summand

Given equation: $\sum_{r = 0}^{10} (\frac{1 0 ^{r + 1} - 1}{1 0 ^{r}}) \cdot^{11} C_{r + 1} = \frac{α ^{11} - 1 1 ^{11}}{1 0 ^{10}}$
Let's simplify the fraction inside the summation first.
$\frac{1 0 ^{r + 1} - 1}{1 0 ^{r}} = \frac{1 0 ^{r + 1}}{1 0 ^{r}} - \frac{1}{1 0 ^{r}}$
Using laws of indices: $10 - 1 0^{- r}$

Distribute the Binomial Coefficient

Substitute the simplified term back into the summation:
$\sum_{r = 0}^{10} (10 - 1 0^{- r}) \cdot^{11} C_{r + 1}$
Distribute $^{11} C_{r + 1}$ inside the bracket:
$\sum_{r = 0}^{10} [10 \cdot^{11} C_{r + 1} - 1 0^{- r} \cdot^{11} C_{r + 1}]$

Split the Summation

Split the summation into two parts using linearity:
$10 \sum_{r = 0}^{10}^{11} C_{r + 1} - \sum_{r = 0}^{10} 1 0^{- r} \cdot^{11} C_{r + 1}$
Let's call the first sum $S_{1}$ and the second sum $S_{2}$ .
LHS $= 10 S_{1} - S_{2}$

Evaluate the First Summation $S_{1}$

Focus on the first sum: $S_{1} = \sum_{r = 0}^{10}^{11} C_{r + 1}$
Expanding the terms by putting $r = 0, 1, \dots, 10$ :
$S_{1} =^{11} C_{1} +^{11} C_{2} +^{11} C_{3} + \dots +^{11} C_{11}$

Apply Binomial Sum Property

Recall the property: $\sum_{k = 0}^{n}^{n} C_{k} = 2^{n}$
For $n = 11$ : $^{11} C_{0} +^{11} C_{1} + \dots +^{11} C_{11} = 2^{11}$
Therefore, $S_{1} = 2^{11} -^{11} C_{0} = 2^{11} - 1$
So, the first part is $10 (2^{11} - 1)$ .

Analyze the Second Summation $S_{2}$

Focus on the second sum: $S_{2} = \sum_{r = 0}^{10} 1 0^{- r} \cdot^{11} C_{r + 1}$
We need the power of $10$ to match the lower index of $C$ .
Rewrite $1 0^{- r}$ as $10 \cdot 1 0^{- (r + 1)}$ :
$S_{2} = 10 \sum_{r = 0}^{10}^{11} C_{r + 1} (\frac{1}{10})^{r + 1}$

Identify the Binomial Expansion

Let $k = r + 1$ . As $r$ goes from $0$ to $10$ , $k$ goes from $1$ to $11$ .
$S_{2} = 10 \sum_{k = 1}^{11}^{11} C_{k} (\frac{1}{10})^{k}$
Recall: $(1 + x)^{n} = \sum_{k = 0}^{n}^{n} C_{k} x^{k}$
The sum is exactly $(1 + \frac{1}{10})^{11} -^{11} C_{0} (\frac{1}{10})^{0}$ .

Calculate the Second Sum Value

Evaluate the expression for $S_{2}$ :
$S_{2} = 10 [(1 + \frac{1}{10})^{11} - 1]$
Simplify the fraction: $1 + \frac{1}{10} = \frac{11}{10}$
$S_{2} = 10 [(\frac{11}{10})^{11} - 1]$

Combine Both Parts

Substitute $S_{1}$ and $S_{2}$ back into LHS $= 10 S_{1} - S_{2}$ :
LHS $= 10 (2^{11} - 1) - 10 [(\frac{11}{10})^{11} - 1]$
Expand the brackets:
LHS $= 10 \cdot 2^{11} - 10 - 10 \cdot (\frac{11}{10})^{11} + 10$

Simplify the Combined Expression

Cancel the constant terms $(- 10 + 10 = 0)$ :
LHS $= 10 \cdot 2^{11} - 10 \cdot \frac{1 1 ^{11}}{1 0 ^{11}}$
Simplify the second term: $\frac{10 \cdot 1 1 ^{11}}{1 0 ^{11}} = \frac{1 1 ^{11}}{1 0 ^{10}}$
LHS $= 10 \cdot 2^{11} - \frac{1 1 ^{11}}{1 0 ^{10}}$

Prepare for Comparison

To get a common denominator of $1 0^{10}$ , multiply the first term by $\frac{1 0 ^{10}}{1 0 ^{10}}$ :
LHS $= \frac{10 \cdot 2 ^{11} \cdot 1 0 ^{10}}{1 0 ^{10}} - \frac{1 1 ^{11}}{1 0 ^{10}}$
Combine the powers of $10$ : $1 0^{1} \cdot 1 0^{10} = 1 0^{11}$
LHS $= \frac{1 0 ^{11} \cdot 2 ^{11} - 1 1 ^{11}}{1 0 ^{10}}$

Final Simplification of LHS

Apply the property $a^{n} \cdot b^{n} = (ab)^{n}$ :
$1 0^{11} \cdot 2^{11} = (10 \cdot 2)^{11} = 2 0^{11}$
Final simplified LHS:
LHS $= \frac{2 0 ^{11} - 1 1 ^{11}}{1 0 ^{10}}$

Compare with RHS and Find $α$

Compare the simplified LHS with the given RHS:
$\frac{2 0 ^{11} - 1 1 ^{11}}{1 0 ^{10}} = \frac{α ^{11} - 1 1 ^{11}}{1 0 ^{10}}$
By direct comparison of the numerators:
$α^{11} = 2 0^{11} ⟹ α = 20$
The correct option is 20.

00:00 / 00:00

The Sigma Insight: Properties of Binomial Coefficients

Analyzing the Setup

The given expression is:

r = 0 \sum 10 (\frac{1 0 ^{r + 1} - 1}{1 0 ^{r}}) \cdot^{11} C_{r + 1} = \frac{α ^{11} - 1 1 ^{11}}{1 0 ^{10}}

To simplify the fraction inside the summation, we apply the laws of indices:

\frac{1 0 ^{r + 1} - 1}{1 0 ^{r}} = \frac{1 0 ^{r + 1}}{1 0 ^{r}} - \frac{1}{1 0 ^{r}} = 10 - 1 0^{- r}

This transformation reduces the complex term into a manageable difference of two components.

The Power of Linearity

Substituting this back into the summation, we obtain:

r = 0 \sum 10 (10 - 1 0^{- r}) \cdot^{11} C_{r + 1}

By distributing the binomial coefficient

^{11} C_{r + 1}

and applying the principle of linearity, we split the expression into two distinct summations,

S_{1}

and

S_{2}

10 r = 0 \sum 10^{11} C_{r + 1} - r = 0 \sum 10 1 0^{- r} \cdot^{11} C_{r + 1}

Conquering the Summits

For the first summation,

S_{1} = \sum_{r = 0}^{10}^{11} C_{r + 1}

, we expand the terms:

S_{1} =^{11} C_{1} +^{11} C_{2} + \dots +^{11} C_{11}

Since the total sum of binomial coefficients is

2^{11}

and we are missing the

^{11} C_{0}

term, we find:

S_{1} = 2^{11} -^{11} C_{0} = 2^{11} - 1

For the second summation,

S_{2} = \sum_{r = 0}^{10} 1 0^{- r} \cdot^{11} C_{r + 1}

, we rewrite

1 0^{- r}

10 \cdot (\frac{1}{10})^{r + 1}

S_{2} = 10 r = 0 \sum 10^{11} C_{r + 1} (\frac{1}{10})^{r + 1}

This represents the binomial expansion of

(1 + \frac{1}{10})^{11}

excluding the first term

^{11} C_{0} (\frac{1}{10})^{0}

S_{2} = 10 [(\frac{11}{10})^{11} - 1]

The Elegant Collapse

We now combine our results to solve for the Left Hand Side (LHS):

L H S = 10 S_{1} - S_{2} = 10 (2^{11} - 1) - 10 [(\frac{11}{10})^{11} - 1]

Expanding this, the constant terms

- 10

and

+ 10

cancel out:

L H S = 10 \cdot 2^{11} - 10 \cdot \frac{1 1 ^{11}}{1 0 ^{11}} = 10 \cdot 2^{11} - \frac{1 1 ^{11}}{1 0 ^{10}}

To match the denominator of the Right Hand Side (RHS), we multiply the first term by

\frac{1 0 ^{10}}{1 0 ^{10}}

L H S = \frac{1 0 ^{11} \cdot 2 ^{11} - 1 1 ^{11}}{1 0 ^{10}} = \frac{2 0 ^{11} - 1 1 ^{11}}{1 0 ^{10}}

Comparing this result to the RHS, we conclude that $α = 20$ .

Common Traps & FAQs

Wait, why do we subtract

^{11} C_{0}

when evaluating the sums?

That is a fantastic observation! The binomial theorem states

\sum_{k = 0}^{n}^{n} C_{k} = 2^{n}

. This sum includes the term for

k = 0

, which is

^{n} C_{0}

. However, our summation starts from

r = 0

, which corresponds to the term

^{11} C_{0 + 1} =^{11} C_{1}

. Since the term

^{11} C_{0}

is missing from our series, we must subtract it from the total sum

2^{11}

to get the correct value.

Why can't we just use energy conservation or some other physics trick here?

I love the enthusiasm! But remember, this is a pure algebraic problem, not a physics one. We are dealing with the structure of binomial coefficients. The 'trick' here is not energy conservation, but the 'linearity of summation' and the 'binomial expansion'—these are the conservation laws of algebra!

How do I know when to split the summation?

Whenever you see a complex expression inside a summation, especially one that looks like a fraction or a product, your first instinct should be to simplify it into additive terms. Linearity allows us to distribute the summation sign over addition and subtraction. If you can break a complex beast into two smaller, familiar ones, you have already won half the battle.

Animated Solution for Mathematics - Binomial Theorem: 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 :

Select Answer:

Visualized Solution

Analyze and Simplify the Summand

Distribute the Binomial Coefficient

Split the Summation

Evaluate the First Summation $S_{1}$

Apply Binomial Sum Property

Analyze the Second Summation $S_{2}$

Identify the Binomial Expansion

Calculate the Second Sum Value

Combine Both Parts

Simplify the Combined Expression

Prepare for Comparison

Final Simplification of LHS

Compare with RHS and Find $α$

The Sigma Insight: Properties of Binomial Coefficients

Analyzing the Setup

The Power of Linearity

Conquering the Summits

The Elegant Collapse

Similar Questions

The value of $\sum_{r = 0}^{6} (^{6} C_{r} \cdot^{6} C_{6 - r})$ is equal to :

Suppose $\sum_{r = 0}^{2023} r^{2} \cdot^{2023} C_{r} = 2023 \times α \times 2^{2022}$ . Then the value of $α$ is ______.

If $\sum_{r = 1}^{30} \frac{r ^{2} ( ^{30} C _{r} ) ^{2}}{^{30} C _{r - 1}} = α \times 2^{29}$ , then $α$ is equal to

If $\sum_{k = 1}^{31} (^{31} C_{k}) (^{31} C_{k - 1}) - \sum_{k = 1}^{30} (^{30} C_{k}) (^{30} C_{k - 1}) = \frac{α ( 60 !)}{( 30 !) ( 31 !)}$ , where $α \in R$ , then the value of $16 α$ is equal to

If $(^{30} C_{0})^{2} + 2 (^{30} C_{1})^{2} + 3 (^{30} C_{2})^{2} + \dots + 31 (^{30} C_{30})^{2} = \frac{α 60 !}{( 30 ! ) ^{2}}$ , then $α$ is equal to

If $(\frac{1}{^{15} C _{0}} + \frac{1}{^{15} C _{1}}) (\frac{1}{^{15} C _{1}} + \frac{1}{^{15} C _{2}}) \dots (\frac{1}{^{15} C _{12}} + \frac{1}{^{15} C _{13}}) = \frac{α ^{13}}{^{14} C _{0} \cdot ^{14} C _{1} \dots ^{14} C _{12}}$ , then $30 α$ is equal to .........

The value $\sum_{r = 0}^{22}^{22} C_{r} \cdot^{23} C_{r}$ is

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

The value of $r$ for which $^{20} C_{r} \cdot^{20} C_{0} +^{20} C_{r - 1} \cdot^{20} C_{1} + \dots +^{20} C_{0} \cdot^{20} C_{r}$ is maximum, is

Let $α = \sum_{k = 0}^{n} \frac{( ^{n} C _{k} ) ^{2}}{k + 1}$ and $β = \sum_{k = 0}^{n - 1} \frac{^{n} C _{k}^{n} C _{k + 1}}{k + 2}$ . If $5 α = 6 β$ , then $n$ equals

Animated Solution for Mathematics - Binomial Theorem: .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 :

Select Answer:

Visualized Solution

Analyze and Simplify the Summand

Distribute the Binomial Coefficient

Split the Summation

Evaluate the First Summation .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } S1​

Apply Binomial Sum Property

Analyze the Second Summation .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } S2​

Identify the Binomial Expansion

Calculate the Second Sum Value

Combine Both Parts

Simplify the Combined Expression

Prepare for Comparison

Final Simplification of LHS

Compare with RHS and Find .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } α

The Sigma Insight: Properties of Binomial Coefficients

Analyzing the Setup

The Power of Linearity

Conquering the Summits

The Elegant Collapse

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The value of ∑r=06​(6Cr​⋅6C6−r​) is equal to :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Suppose ∑r=02023​r2⋅2023Cr​=2023×α×22022. Then the value of α is ______.

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If ∑r=130​30Cr−1​r2(30Cr​)2​=α×229, then α is equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If ∑k=131​(31Ck​)(31Ck−1​)−∑k=130​(30Ck​)(30Ck−1​)=(30!)(31!)α(60!)​, where α∈R, then the value of 16α is equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If (30C0​)2+2(30C1​)2+3(30C2​)2+⋯+31(30C30​)2=(30!)2α60!​, then α is equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If (15C0​1​+15C1​1​)(15C1​1​+15C2​1​)⋯(15C12​1​+15C13​1​)=14C0​⋅14C1​⋯14C12​α13​, then 30α is equal to .........

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The value ∑r=022​22Cr​⋅23Cr​ is

.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 :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The value of r for which 20Cr​⋅20C0​+20Cr−1​⋅20C1​+⋯+20C0​⋅20Cr​ is maximum, is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let α=∑k=0n​k+1(nCk​)2​ and β=∑k=0n−1​k+2nCkn​Ck+1​​. If 5α=6β, then n equals

Animated Solution for Mathematics - Binomial Theorem: 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 :

Evaluate the First Summation $S_{1}$

Analyze the Second Summation $S_{2}$

Compare with RHS and Find $α$

The value of $\sum_{r = 0}^{6} (^{6} C_{r} \cdot^{6} C_{6 - r})$ is equal to :

Suppose $\sum_{r = 0}^{2023} r^{2} \cdot^{2023} C_{r} = 2023 \times α \times 2^{2022}$ . Then the value of $α$ is ______.

If $\sum_{r = 1}^{30} \frac{r ^{2} ( ^{30} C _{r} ) ^{2}}{^{30} C _{r - 1}} = α \times 2^{29}$ , then $α$ is equal to

If $\sum_{k = 1}^{31} (^{31} C_{k}) (^{31} C_{k - 1}) - \sum_{k = 1}^{30} (^{30} C_{k}) (^{30} C_{k - 1}) = \frac{α ( 60 !)}{( 30 !) ( 31 !)}$ , where $α \in R$ , then the value of $16 α$ is equal to

If $(^{30} C_{0})^{2} + 2 (^{30} C_{1})^{2} + 3 (^{30} C_{2})^{2} + \dots + 31 (^{30} C_{30})^{2} = \frac{α 60 !}{( 30 ! ) ^{2}}$ , then $α$ is equal to

If $(\frac{1}{^{15} C _{0}} + \frac{1}{^{15} C _{1}}) (\frac{1}{^{15} C _{1}} + \frac{1}{^{15} C _{2}}) \dots (\frac{1}{^{15} C _{12}} + \frac{1}{^{15} C _{13}}) = \frac{α ^{13}}{^{14} C _{0} \cdot ^{14} C _{1} \dots ^{14} C _{12}}$ , then $30 α$ is equal to .........

The value $\sum_{r = 0}^{22}^{22} C_{r} \cdot^{23} C_{r}$ is

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

The value of $r$ for which $^{20} C_{r} \cdot^{20} C_{0} +^{20} C_{r - 1} \cdot^{20} C_{1} + \dots +^{20} C_{0} \cdot^{20} C_{r}$ is maximum, is

Let $α = \sum_{k = 0}^{n} \frac{( ^{n} C _{k} ) ^{2}}{k + 1}$ and $β = \sum_{k = 0}^{n - 1} \frac{^{n} C _{k}^{n} C _{k + 1}}{k + 2}$ . If $5 α = 6 β$ , then $n$ equals