Why is when the question says coefficient of ?

Because by the symmetry of binomial coefficients, , so the coefficient of is .

JEE MainMathematics

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Binomial Theorem: JEE Main Mathematics Question with Solution

Q: Why does reversing the index work?

Summing terms in reverse order yields the exact same total, but makes the quadratic term monomial in .

a_r

is the coefficient of

x^{10-r}

in the Binomial expansion of

(1+x)^{10}

, then

\sum_{r=1}^{10} r^3 \left(\frac{a_r}{a_{r-1}}\right)^2

is equal to

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A5445B3025C4895D1210

Step-by-step solutionView

Correct answer

The coefficient $a_r = \binom{10}{r}$ , giving the ratio $\frac{a_r}{a_{r-1}} = \frac{11-r}{r}$ ; summing $r^3 \left(\frac{11-r}{r}\right)^2 = r(11-r)^2$ from $r=1$ to $10$ yields 1210.

Option analysis

Why each option works or fails

A · 5445

Adding the squared terms without factoring in the surviving $r$ after cancellation, evaluating $\sum (11-r)^2$ or miscalculating $\sum r^3$ directly. Cancel $r^2$ from the denominator against $r^3$ to leave $r(11-r)^2$ , then expand into standard polynomial power sums.

B · 3025

Forgetting the ratio simplification and computing $\left(\sum_{r=1}^{10} r\right)^2 = 55^2 = 3025$ . Simplify the summand algebraically first before applying any summation formulas.

C · 4895

Flipping the ratio identity to $\frac{a_r}{a_{r-1}} = \frac{r}{11-r}$ and incorrectly expanding the resulting algebra. Recall that $\frac{\binom{n}{r}}{\binom{n}{r-1}} = \frac{n-r+1}{r}$ , so for $n=10$ , the numerator is $11-r$ and the denominator is $r$ .

D · 1210

None. The simplification $r(11-r)^2$ and subsequent summation correctly yield 1210. Correctly simplified $r^3\left(\frac{11-r}{r}\right)^2 = r(11-r)^2$ and evaluated the sum to get 1210.

Reviewed route

Solution

StepWorking

01given

In the expansion of $(1+x)^{10}$ , the general term is $T_{k+1} = \binom{10}{k} x^k$ . We are given that $a_r$ is the coefficient of $x^{10-r}$ .

02approach

Find $a_r$ explicitly using symmetry of binomial coefficients, evaluate the ratio $\frac{a_r}{a_{r-1}}$ , substitute it into the given summation, simplify the general term into polynomial powers of $r$ , and apply standard power sum formulas $\sum r$ , $\sum r^2$ , $\sum r^3$ .

03execute

Since $a_r$ is the coefficient of $x^{10-r}$ , $a_r = \binom{10}{10-r} = \binom{10}{r}$ . Similarly, $a_{r-1} = \binom{10}{r-1}$ . The ratio is: $\frac{a_r}{a_{r-1}} = \frac{\binom{10}{r}}{\binom{10}{r-1}} = \frac{10 - r + 1}{r} = \frac{11 - r}{r}$

04execute

Substitute the ratio into the summation: $S = \sum_{r=1}^{10} r^3 \left(\frac{11-r}{r}\right)^2 = \sum_{r=1}^{10} r (11-r)^2 = \sum_{r=1}^{10} r (r^2 - 22r + 121) = \sum_{r=1}^{10} (r^3 - 22r^2 + 121r)$

05execute

Apply the standard power summation formulas for $n = 10$ : $\sum_{r=1}^{10} r = \frac{10 \times 11}{2} = 55$ $\sum_{r=1}^{10} r^2 = \frac{10 \times 11 \times 21}{6} = 385$ $\sum_{r=1}^{10} r^3 = \left(\frac{10 \times 11}{2}\right)^2 = 55^2 = 3025$ Now compute $S$ : $S = 3025 - 22(385) + 121(55) = 3025 - 8470 + 6655 = 1210$

✓verify

Check factoring out common terms: $S = \sum_{r=1}^{10} r(11-r)^2$ . Substituting $k = 11-r$ gives $\sum_{k=1}^{10} (11-k)k^2 = 11\sum k^2 - \sum k^3 = 11(385) - 3025 = 4235 - 3025 = 1210$ . The two methods match exactly.

✓ Source and academic review

Question type: Single correct
Exam relevance: JEE Main · Mathematics
Concepts assessed: Mathematics
Academic status: Reviewed by official_key
Source: pyq
Editorial review: 8 September 2026

Quick checks

Students also ask

Why is

a_r = \binom{10}{r}

when the question says coefficient of

x^{10-r}

Because by the symmetry of binomial coefficients, $\binom{n}{n-r} = \binom{n}{r}$ , so the coefficient of $x^{10-r}$ is $\binom{10}{10-r} = \binom{10}{r}$ .

Why does reversing the index work?

Summing terms in reverse order $r = 10, 9, \dots, 1$ yields the exact same total, but makes the quadratic term monomial in $k$ .

Answer

The coefficient $a_r = \binom{10}{r}$ , giving the ratio $\frac{a_r}{a_{r-1}} = \frac{11-r}{r}$ ; summing $r^3 \left(\frac{11-r}{r}\right)^2 = r(11-r)^2$ from $r=1$ to $10$ yields 1210.

Why each option works or fails

A: 5445 - Adding the squared terms without factoring in the surviving $r$ after cancellation, evaluating $\sum (11-r)^2$ or miscalculating $\sum r^3$ directly. Cancel $a_r = \binom{10}{r}$ from the denominator against $r^3$ to leave $r(11-r)^2$ , then expand into standard polynomial power sums.
B: 3025 - Forgetting the ratio simplification and computing $\left(\sum_{r=1}^{10} r\right)^2 = 55^2 = 3025$ . Simplify the summand algebraically first before applying any summation formulas.
C: 4895 - Flipping the ratio identity to $\frac{a_r}{a_{r-1}} = \frac{r}{11-r}$ and incorrectly expanding the resulting algebra. Recall that $\frac{\binom{n}{r}}{\binom{n}{r-1}} = \frac{n-r+1}{r}$ , so for $n=10$ , the numerator is $11-r$ and the denominator is $r$ .
D · correct: 1210 - None. The simplification $r(11-r)^2$ and subsequent summation correctly yield 1210. Correctly simplified $r^3\left(\frac{11-r}{r}\right)^2 = r(11-r)^2$ and evaluated the sum to get 1210.

Step-by-step solution

given: In the expansion of $(1+x)^{10}$ , the general term is $T_{k+1} = \binom{10}{k} x^k$ . We are given that $a_r$ is the coefficient of $x^{10-r}$ .
approach: Find $a_r$ explicitly using symmetry of binomial coefficients, evaluate the ratio $\frac{a_r}{a_{r-1}}$ , substitute it into the given summation, simplify the general term into polynomial powers of $r$ , and apply standard power sum formulas $\sum r$ , $\sum r^2$ , $\sum r^3$ .
execute: Since $a_r$ is the coefficient of $x^{10-r}$ , $a_r = \binom{10}{10-r} = \binom{10}{r}$ . Similarly, $a_{r-1} = \binom{10}{r-1}$ . The ratio is: $\frac{a_r}{a_{r-1}} = \frac{\binom{10}{r}}{\binom{10}{r-1}} = \frac{10 - r + 1}{r} = \frac{11 - r}{r}$
execute: Substitute the ratio into the summation: $S = \sum_{r=1}^{10} r^3 \left(\frac{11-r}{r}\right)^2 = \sum_{r=1}^{10} r (11-r)^2 = \sum_{r=1}^{10} r (r^2 - 22r + 121) = \sum_{r=1}^{10} (r^3 - 22r^2 + 121r)$
execute: Apply the standard power summation formulas for $n = 10$ : $\sum_{r=1}^{10} r = \frac{10 \times 11}{2} = 55$ $\sum_{r=1}^{10} r^2 = \frac{10 \times 11 \times 21}{6} = 385$ $\sum_{r=1}^{10} r^3 = \left(\frac{10 \times 11}{2}\right)^2 = 55^2 = 3025$ Now compute $S$ : $S = 3025 - 22(385) + 121(55) = 3025 - 8470 + 6655 = 1210$
verify: Check factoring out common terms: $S = \sum_{r=1}^{10} r(11-r)^2$ . Substituting $k = 11-r$ gives $\sum_{k=1}^{10} (11-k)k^2 = 11\sum k^2 - \sum k^3 = 11(385) - 3025 = 4235 - 3025 = 1210$ . The two methods match exactly.

Shortcut:

When to use it: Whenever the summand contains symmetric factors like $r(n+1-r)^2$ , substituting $k = n+1-r$ halves the calculation.

given: From $\frac{a_r}{a_{r-1}} = \frac{11-r}{r}$ , the sum is $S = \sum_{r=1}^{10} r(11-r)^2$ .

approach: Substitute $k = 11 - r$ . As $r$ goes from $1$ to $10$ , $k$ also goes from $10$ down to $1$ . This immediately simplifies the summand to $(11-k)k^2 = 11k^2 - k^3$ .

execute: Evaluate the transformed sum: $S = \sum_{k=1}^{10} (11k^2 - k^3) = 11 \sum_{k=1}^{10} k^2 - \sum_{k=1}^{10} k^3$ $S = 11(385) - 3025 = 4235 - 3025 = 1210$

verify: Factoring 55 out: $55 \times (11 \times 7 - 55) = 55 \times (77 - 55) = 55 \times 22 = 1210$ .

Binomial Theorem: JEE Main Mathematics Question with Solution

The coefficient ar=(10r)a_r = \binom{10}{r}ar​=(r10​), giving the ratio arar−1=11−rr\frac{a_r}{a_{r-1}} = \frac{11-r}{r}ar−1​ar​​=r11−r​; summing r3(11−rr)2=r(11−r)2r^3 \left(\frac{11-r}{r}\right)^2 = r(11-r)^2r3(r11−r​)2=r(11−r)2 from r=1r=1r=1 to 101010 yields 1210.

01given

02approach

03execute

04execute

05execute

✓verify

Students also ask

The coefficient $a_r = \binom{10}{r}$ , giving the ratio $\frac{a_r}{a_{r-1}} = \frac{11-r}{r}$ ; summing $r^3 \left(\frac{11-r}{r}\right)^2 = r(11-r)^2$ from $r=1$ to $10$ yields 1210.