JEE MainMathematics

Answer verified against the official NTA key

+4 marks−1 if incorrectSingle correctPrevious-year question

Sum of a Binomial Ratio Series

Q: Why is a_r = binom10r when the question says coefficient of x^(10-r)?

Because by the symmetry of binomial coefficients, binomnn-r = binomnr, so the coefficient of x^(10-r) is binom1010-r = binom10r.

Q: Why does reversing the index work?

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

If a_r is the coefficient of x^(10-r) in the Binomial expansion of (1+x)^10, then sum from r=1 to 10 of r^3 ((a_r)/(a_(r-1)))^2 is equal to

Your answer stays private

What feels right?

No score. Commit to your first instinct. We’ll show what your mind noticed and what it missed.

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 $r^{3} {(\frac{11 - r}{r})}^{2} = r (11 - r)^{2}$ 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 ${(\sum_{r = 1}^{10} r)}^{2} = 55^{2} = 3025$ . 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

The simplification $r (11 - r)^{2}$ and subsequent summation correctly yield 1210. Correctly simplified $r^{3} {(\frac{11 - r}{r})}^{2} = r (11 - r)^{2}$ 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: $S = \sum_{r = 1}^{10} r^{3} {(\frac{11 - r}{r})}^{2} = \sum_{r = 1}^{10} r (11 - r)^{2} = \sum_{r = 1}^{10} r (r^{2} - 22 r + 121) = \sum_{r = 1}^{10} (r^{3} - 22 r^{2} + 121 r)$

05Execute

Apply the standard power summation formulas for $n = 10$ : $\sum_{r = 1}^{10} r = \frac{10 \times 11}{2} = 55$ $\sum_{r = 1}^{10} r^{2} = \frac{10 \times 11 \times 21}{6} = 385$ $\sum_{r = 1}^{10} r^{3} = {(\frac{10 \times 11}{2})}^{2} = 55^{2} = 3025$ 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
Academic status: Answer verified against the official NTA key
Source: Previous-year question
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 $r^{3} {(\frac{11 - r}{r})}^{2} = r (11 - r)^{2}$ 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 $r^{2}$ 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 ${(\sum_{r = 1}^{10} r)}^{2} = 55^{2} = 3025$ . 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 - The simplification $r (11 - r)^{2}$ and subsequent summation correctly yield 1210. Correctly simplified $r^{3} {(\frac{11 - r}{r})}^{2} = r (11 - r)^{2}$ and evaluated the sum to get 1210.

Step-by-step solution

Given: In the expansion of $(1 + x)^{10}$ , the general term is $r (11 - r)^{2}$ . 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: $S = \sum_{r = 1}^{10} r^{3} {(\frac{11 - r}{r})}^{2} = \sum_{r = 1}^{10} r (11 - r)^{2} = \sum_{r = 1}^{10} r (r^{2} - 22 r + 121) = \sum_{r = 1}^{10} (r^{3} - 22 r^{2} + 121 r)$
Execute: Apply the standard power summation formulas for $n = 10$ : $\sum_{r = 1}^{10} r = \frac{10 \times 11}{2} = 55$ $\sum_{r = 1}^{10} r^{2} = \frac{10 \times 11 \times 21}{6} = 385$ $\sum_{r = 1}^{10} r^{3} = {(\frac{10 \times 11}{2})}^{2} = 55^{2} = 3025$ 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 $(11 - k) k^{2} = 11 k^{2} - k^{3}$ .

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

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

Sum of a Binomial Ratio Series

What feels right?

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

Why each option works or fails

Solution

01Given

02Approach

03Execute

04Execute

05Execute

✓Verify

We think you should solve this next.

The coefficient of $x^{- 6}$ , in the expansion of…

If the constant term in the binomial expansion of…

The points of intersection of the line $a x + b y = 0, (a \neq b)$ and the circle…

Students also ask

Sum of a Binomial Ratio Series

The coefficient ar=(10r)ar​=(r10​), giving the ratio arar−1=11−rrar−1​ar​​=r11−r​; summing r3(11−rr)2=r(11−r)2r3(r11−r​)2=r(11−r)2 from r=1r=1 to 1010 yields 1210.

01Given

02Approach

03Execute

04Execute

05Execute

✓Verify

The coefficient of x−6x−6, in the expansion of…

If the constant term in the binomial expansion of…

The points of intersection of the line ax+by=0,(a≠b)ax+by=0,(a=b) and the circle…

Students also ask

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

The coefficient of $x^{- 6}$ , in the expansion of…

The points of intersection of the line $a x + b y = 0, (a \neq b)$ and the circle…