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JEE Mathematics 2023 Question with Solution

If ara_r is the coefficient of x10−rx^{10-r} in the binomial expansion of (1+x)10(1 + x)^{10}, then ∑r=110r3(arar−1)2\sum_{r=1}^{10} r^3 \left( \frac{a_r}{a_{r-1}} \right)^2 is equal to:

  • A

    48954895

  • B

    12101210

  • C

    54455445

  • D

    30253025

Answer

Correct answer:B

Step-by-step solution

Standard Method

Given: ara_r is the coefficient of x10−rx^{10-r} in (1+x)10(1+x)^{10}.

Find: The value of

∑r=110r3(arar−1)2\sum_{r=1}^{10} r^3 \left( \frac{a_r}{a_{r-1}} \right)^2

From the binomial expansion,

ar=(1010−r)=(10r)a_r = \binom{10}{10-r} = \binom{10}{r}

Therefore,

arar−1=(10r)(10r−1)=11−rr\frac{a_r}{a_{r-1}} = \frac{\binom{10}{r}}{\binom{10}{r-1}} = \frac{11-r}{r}

Substituting into the sum,

∑r=110r3(arar−1)2=∑r=110r3(11−rr)2=∑r=110r(11−r)2\sum_{r=1}^{10} r^3 \left( \frac{a_r}{a_{r-1}} \right)^2 = \sum_{r=1}^{10} r^3 \left( \frac{11-r}{r} \right)^2 = \sum_{r=1}^{10} r(11-r)^2

Now expand:

(11−r)2=121−22r+r2(11-r)^2 = 121 - 22r + r^2

So,

∑r=110r(11−r)2=∑r=110(121r−22r2+r3)\sum_{r=1}^{10} r(11-r)^2 = \sum_{r=1}^{10} (121r - 22r^2 + r^3)

Using standard sums,

∑r=110r=55,∑r=110r2=385,∑r=110r3=3025\sum_{r=1}^{10} r = 55, \qquad \sum_{r=1}^{10} r^2 = 385, \qquad \sum_{r=1}^{10} r^3 = 3025

Hence,

121⋅55−22⋅385+3025=6655−8470+3025=1210121 \cdot 55 - 22 \cdot 385 + 3025 = 6655 - 8470 + 3025 = 1210

Therefore, the correct option is B and the value of the sum is 12101210.

Expanded Summation Evaluation

Given: ar=(10r)a_r = \binom{10}{r} for the expansion of (1+x)10(1+x)^{10}.

Find: ∑r=110r3(arar−1)2\sum_{r=1}^{10} r^3 \left( \frac{a_r}{a_{r-1}} \right)^2

First compute the ratio of consecutive coefficients:

arar−1=(10r)(10r−1)=10−r+1r=11−rr\frac{a_r}{a_{r-1}} = \frac{\binom{10}{r}}{\binom{10}{r-1}} = \frac{10-r+1}{r} = \frac{11-r}{r}

Then,

r3(arar−1)2=r3⋅(11−r)2r2=r(11−r)2r^3 \left( \frac{a_r}{a_{r-1}} \right)^2 = r^3 \cdot \frac{(11-r)^2}{r^2} = r(11-r)^2

So the required sum becomes

∑r=110r(11−r)2\sum_{r=1}^{10} r(11-r)^2

Expand the square:

(11−r)2=121−22r+r2(11-r)^2 = 121 - 22r + r^2

Hence,

∑r=110r(11−r)2=∑r=110(121r−22r2+r3)\sum_{r=1}^{10} r(11-r)^2 = \sum_{r=1}^{10} (121r - 22r^2 + r^3)

Now evaluate each part separately:

∑r=110121r=121∑r=110r=121⋅55=6655\sum_{r=1}^{10} 121r = 121 \sum_{r=1}^{10} r = 121 \cdot 55 = 6655 ∑r=11022r2=22∑r=110r2=22⋅385=8470\sum_{r=1}^{10} 22r^2 = 22 \sum_{r=1}^{10} r^2 = 22 \cdot 385 = 8470 ∑r=110r3=(10⋅112)2=552=3025\sum_{r=1}^{10} r^3 = \left( \frac{10 \cdot 11}{2} \right)^2 = 55^2 = 3025

Therefore,

6655−8470+3025=12106655 - 8470 + 3025 = 1210

So the required value is 12101210, hence the correct option is B.

Common mistakes

  • Using ar=(1010−r)a_r = \binom{10}{10-r} and then failing to simplify it to (10r)\binom{10}{r}. These two are equal by symmetry of binomial coefficients. Always use (nk)=(nn−k)\binom{n}{k} = \binom{n}{n-k} before forming the ratio.

  • Writing arar−1=r11−r\frac{a_r}{a_{r-1}} = \frac{r}{11-r} instead of 11−rr\frac{11-r}{r}. This reverses numerator and denominator and changes the entire sum. Compute the ratio carefully from consecutive binomial coefficients.

  • Not cancelling the factor of r2r^2 after substitution into r3(11−rr)2r^3\left(\frac{11-r}{r}\right)^2. The expression simplifies to r(11−r)2r(11-r)^2, not r3(11−r)2r^3(11-r)^2. Simplify powers before expanding the summation.

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