Wait, why is $a_r = \binom{10}{r}$? The problem says $a_r$ is the coefficient of $x^{10-r}$ in $(1+x)^{10}$. Shouldn't it be $\binom{10}{10-r}$?

You are absolutely correct! It is $\binom{10}{10-r}$. However, thanks to the symmetry property of binomial coefficients, $\binom{n}{k} = \binom{n}{n-k}$. So, $\binom{10}{10-r}$ is exactly the same as $\binom{10}{10-(10-r)}$, which simplifies to $\binom{10}{r}$. It's a beautiful property that makes our lives much easier!

JEE Main 2023 (25 January Shift 1)

LEVELJEE Main

Animated Solution for Mathematics - Binomial Theorem: If $a_{r}$ is the coefficient of $x^{10 - r}$ in the Binomial expansion of $(1 + x)^{10}$ , then $\sum_{r = 1}^{10} r^{3} (\frac{a _{r}}{a _{r - 1}})^{2}$ is equal to

Q: Why does the summation start from $r=1$? What happens to $r=0$?

Great observation! If $r=0$, the term $a_{r-1}$ would become $a_{-1}$, which is undefined in this context. The problem specifically asks for the sum from $r=1$ to $10$, so we only care about the values where the ratio is well-defined. Always check the domain of your summation!

Select Answer:

Visualized Solution

Identify the Coefficient $a_{r}$

Given: $a_{r}$ is the coefficient of $x^{10 - r}$ in $(1 + x)^{10}$ .
We need to find the exact binomial coefficient for this term.

General Term Formula

The general term in $(1 + x)^{n}$ is $T_{k + 1} = (k n) x^{k}$ .
For our term, the power of $x$ is $10 - r$ .
Therefore, $k = 10 - r$ .

Simplify $a_{r}$

The coefficient is $a_{r} = (10 - r 10)$ .
Using the symmetry property: $(k n) = (n - k n)$ .
$a_{r} = (10 - ( 10 - r ) 10) = (r 10)$ .

Setup the Ratio

The target expression contains the ratio $\frac{a _{r}}{a _{r - 1}}$ .
Substitute $a_{r} = (r 10)$ .
$a_{r - 1} = (r - 1 10)$ .
Ratio: $\frac{( r 10 )}{( r - 1 10 )}$ .

Evaluate the Ratio

Standard formula: $\frac{( r n )}{( r - 1 n )} = \frac{n - r + 1}{r}$ .
Here, $n = 10$ .
$\frac{a _{r}}{a _{r - 1}} = \frac{10 - r + 1}{r} = \frac{11 - r}{r}$ .

Substitute into Summation

Target expression: $S = \sum_{r = 1}^{10} r^{3} (\frac{a _{r}}{a _{r - 1}})^{2}$ .
Substitute the simplified ratio:
$S = \sum_{r = 1}^{10} r^{3} (\frac{11 - r}{r})^{2}$ .

Simplify the Expression

Square the fraction: $(\frac{11 - r}{r})^{2} = \frac{( 11 - r ) ^{2}}{r ^{2}}$ .
Substitute back: $S = \sum_{r = 1}^{10} r^{3} \frac{( 11 - r ) ^{2}}{r ^{2}}$ .
Cancel $r^{2}$ from numerator and denominator:
$S = \sum_{r = 1}^{10} r (11 - r)^{2}$ .

Expand the Square

Expand $(11 - r)^{2}$ using $(a - b)^{2} = a^{2} - 2 ab + b^{2}$ .
$(11 - r)^{2} = 121 - 22 r + r^{2}$ .
Substitute into $S$ :
$S = \sum_{r = 1}^{10} r (121 - 22 r + r^{2})$ .

Distribute and Split

Distribute $r$ inside the bracket:
$S = \sum_{r = 1}^{10} (121 r - 22 r^{2} + r^{3})$ .
Use linearity of summation to split into three parts:
$S = 121 \sum_{r = 1}^{10} r - 22 \sum_{r = 1}^{10} r^{2} + \sum_{r = 1}^{10} r^{3}$ .

Standard Summation Formulas

For $n = 10$ :
$\sum_{r = 1}^{n} r = \frac{n ( n + 1 )}{2}$
$\sum_{r = 1}^{n} r^{2} = \frac{n ( n + 1 ) ( 2 n + 1 )}{6}$
$\sum_{r = 1}^{n} r^{3} = [\frac{n ( n + 1 )}{2}]^{2}$

Calculate Standard Sums

$\sum_{r = 1}^{10} r = \frac{10 \times 11}{2} = 55$ .
$\sum_{r = 1}^{10} r^{2} = \frac{10 \times 11 \times 21}{6} = 385$ .
$\sum_{r = 1}^{10} r^{3} = 5 5^{2} = 3025$ .

Substitute Values

Recall: $S = 121 \sum_{r = 1}^{10} r - 22 \sum_{r = 1}^{10} r^{2} + \sum_{r = 1}^{10} r^{3}$ .
Substitute the calculated values:
$S = 121 (55) - 22 (385) + 3025$ .

Final Calculation

Calculate each term:
$121 \times 55 = 6655$ .
$22 \times 385 = 8470$ .
$S = 6655 - 8470 + 3025$ .
$S = 9680 - 8470 = 1210$ .
Final Answer: 1210

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The Sigma Insight: Properties of Binomial Coefficients

Analyzing the Setup

We are given that

a_{r}

is the coefficient of

x^{10 - r}

in the expansion of

(1 + x)^{10}

. The general term in the expansion of

(1 + x)^{n}

is given by

T_{k + 1} = (k n) x^{k}

Here, our

n = 10

, and we are looking for the coefficient of

x^{10 - r}

, so we set

k = 10 - r

. This gives us

a_{r} = (10 - r 10)

By applying the symmetry property

(k n) = (n - k n)

, we find that

(10 - r 10)

transforms into

(r 10)

. This simple substitution is the key to unlocking the entire problem.

The Ratio Shortcut

The summation we need to evaluate is

S = \sum_{r = 1}^{10} r^{3} (\frac{a _{r}}{a _{r - 1}})^{2}

. We have

a_{r} = (r 10)

and

a_{r - 1} = (r - 1 10)

Instead of writing out the factorials, we use the standard JEE ratio formula:

\frac{( r n )}{( r - 1 n )} = \frac{n - r + 1}{r}

Substituting

n = 10

, our ratio becomes:

\frac{10 - r + 1}{r} = \frac{11 - r}{r}

Algebraic Simplification

Now, let's substitute this back into our summation

S

S = r = 1 \sum 10 r^{3} (\frac{11 - r}{r})^{2}

Squaring the fraction, we get

\frac{( 11 - r ) ^{2}}{r ^{2}}

. When we multiply this by

r^{3}

, the

r^{2}

in the denominator cancels out perfectly with

r^{2}

from the

r^{3}

, leaving us with

r (11 - r)^{2}

The expression is now:

S = r = 1 \sum 10 r (11 - r)^{2}

Expanding

(11 - r)^{2}

gives us

121 - 22 r + r^{2}

. Multiplying by

r

, we get

121 r - 22 r^{2} + r^{3}

The Final Calculation

Because summation is a linear operator, we can split this into three manageable parts:

S = 121 r = 1 \sum 10 r - 22 r = 1 \sum 10 r^{2} + r = 1 \sum 10 r^{3}

Using the standard formulas for

n = 10

r = 1 \sum 10 r = 55, r = 1 \sum 10 r^{2} = 385, r = 1 \sum 10 r^{3} = 3025

Substituting these values:

S = 121 (55) - 22 (385) + 3025

S = 6655 - 8470 + 3025

S = 1210

Through symmetry, ratio properties, and careful algebraic simplification, we have arrived at the final answer: 1210.

Common Traps & FAQs

Wait, why is

a_{r} = (r 10)

? The problem says

a_{r}

is the coefficient of

x^{10 - r}

(1 + x)^{10}

. Shouldn't it be

(10 - r 10)

You are absolutely correct! It is

(10 - r 10)

. However, thanks to the symmetry property of binomial coefficients,

(k n) = (n - k n)

. So,

(10 - r 10)

is exactly the same as

(10 - ( 10 - r ) 10)

, which simplifies to

(r 10)

. It's a beautiful property that makes our lives much easier!

Can I just expand the factorials for the ratio

\frac{a _{r}}{a _{r - 1}}

? It seems safer.

While expanding factorials is technically correct, it is a dangerous trap in JEE Advanced. It consumes precious time and increases the probability of a calculation error. The ratio formula

\frac{( r n )}{( r - 1 n )} = \frac{n - r + 1}{r}

is a standard, high-yield shortcut. Master it, and you'll save minutes that you can spend on tougher problems!

Why does the summation start from

r = 1

? What happens to

r = 0

Great observation! If

r = 0

, the term

a_{r - 1}

would become

a_{- 1}

, which is undefined in this context. The problem specifically asks for the sum from

r = 1

10

, so we only care about the values where the ratio is well-defined. Always check the domain of your summation!

Animated Solution for Mathematics - Binomial Theorem: If $a_{r}$ is the coefficient of $x^{10 - r}$ in the Binomial expansion of $(1 + x)^{10}$ , then $\sum_{r = 1}^{10} r^{3} (\frac{a _{r}}{a _{r - 1}})^{2}$ is equal to

Select Answer:

Visualized Solution

Identify the Coefficient $a_{r}$

General Term Formula

Simplify $a_{r}$

Setup the Ratio

Evaluate the Ratio

Substitute into Summation

Simplify the Expression

Expand the Square

Distribute and Split

Standard Summation Formulas

Calculate Standard Sums

Substitute Values

Final Calculation

The Sigma Insight: Properties of Binomial Coefficients

Analyzing the Setup

The Ratio Shortcut

Algebraic Simplification

The Final Calculation

Similar Questions

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

The sum, of the coefficients of the first 50 terms in the binomial expansion of $(1 - x)^{100}$ , is equal to

The sum of the coefficients of three consecutive terms in the binomial expansion of $(1 + x)^{n + 2}$ , which are in the ratio $1 : 3 : 5$ , 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 $^{20} C_{r}$ is the co-efficient of $x^{r}$ in the expansion of $(1 + x)^{20}$ , then the value of $\sum_{r = 0}^{20} r^{2}^{20} C_{r}$ is equal to :

If for some positive integer $n$ , the coefficients of three consecutive terms in the binomial expansion of $(1 + x)^{n + 5}$ are in the ratio $5 : 10 : 14$ , then the largest coefficient in this expansion is :

If the sum of the coefficients in the expansion of $(a + b)^{n}$ is 4096, then the greatest coefficient in the expansion is

If $A$ denotes the sum of all the coefficients in the expansion of $(1 - 3 x + 10 x^{2})^{n}$ and $B$ denotes the sum of all the coefficients in the expansion of $(1 + x^{2})^{n}$ , then :

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 .

Select Answer:

Visualized Solution

Identify the Coefficient .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ar​

General Term Formula

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

Setup the Ratio

Evaluate the Ratio

Substitute into Summation

Simplify the Expression

Expand the Square

Distribute and Split

Standard Summation Formulas

Calculate Standard Sums

Substitute Values

Final Calculation

The Sigma Insight: Properties of Binomial Coefficients

Analyzing the Setup

The Ratio Shortcut

Algebraic Simplification

The Final Calculation

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The sum, of the coefficients of the first 50 terms in the binomial expansion of (1−x)100, is equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The sum of the coefficients of three consecutive terms in the binomial expansion of (1+x)n+2, which are in the ratio 1:3:5, 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 20Cr​ is the co-efficient of xr in the expansion of (1+x)20, then the value of ∑r=020​r220Cr​ 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 in the expansion of (a+b)n is 4096, then the greatest coefficient in the expansion is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If A denotes the sum of all the coefficients in the expansion of (1−3x+10x2)n and B denotes the sum of all the coefficients in the expansion of (1+x2)n, then :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let (1+x+x2)10=a0​+a1​x+a2​x2+⋯+a20​x20. If (a1​+a3​+a5​+⋯+a19​)−11a2​=121k, then k is equal to .

Identify the Coefficient $a_{r}$

Simplify $a_{r}$

The sum, of the coefficients of the first 50 terms in the binomial expansion of $(1 - x)^{100}$ , is equal to

The sum of the coefficients of three consecutive terms in the binomial expansion of $(1 + x)^{n + 2}$ , which are in the ratio $1 : 3 : 5$ , 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 $^{20} C_{r}$ is the co-efficient of $x^{r}$ in the expansion of $(1 + x)^{20}$ , then the value of $\sum_{r = 0}^{20} r^{2}^{20} C_{r}$ is equal to :

If the sum of the coefficients in the expansion of $(a + b)^{n}$ is 4096, then the greatest coefficient in the expansion is

If $A$ denotes the sum of all the coefficients in the expansion of $(1 - 3 x + 10 x^{2})^{n}$ and $B$ denotes the sum of all the coefficients in the expansion of $(1 + x^{2})^{n}$ , then :

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 .