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NVAMediumJEE 2025Simple Applications

JEE Mathematics 2025 Question with Solution

The sum of the series 2×1×20C4−3×2×20C5+4×3×20C6−5×4×20C7+⋯+18×17×20C202 \times 1 \times {}^{20}C_4 - 3 \times 2 \times {}^{20}C_5 + 4 \times 3 \times {}^{20}C_6 - 5 \times 4 \times {}^{20}C_7 + \dots + 18 \times 17 \times {}^{20}C_{20}, is equal to

Answer

Correct answer:34

Step-by-step solution

Standard Method

Given:

S=∑k=420(−1)k(k−2)(k−3)(20k)S=\sum_{k=4}^{20}(-1)^k (k-2)(k-3){20\choose k}

Find: The value of SS.

Write the given series in summation form:

S=∑k=420(−1)k(k−2)(k−3)(20k)S=\sum_{k=4}^{20}(-1)^k (k-2)(k-3){20\choose k}

Now expand

(k−2)(k−3)=k(k−1)−4k+6(k-2)(k-3)=k(k-1)-4k+6

So,

S=∑k=420(−1)kk(k−1)(20k)⏟A−4∑k=420(−1)kk(20k)⏟B+6∑k=420(−1)k(20k)⏟CS=\underbrace{\sum_{k=4}^{20}(-1)^k k(k-1){20\choose k}}_{A}-4\underbrace{\sum_{k=4}^{20}(-1)^k k{20\choose k}}_{B}+6\underbrace{\sum_{k=4}^{20}(-1)^k {20\choose k}}_{C}

Now compute AA using the identity

k(k−1)(nk)=n(n−1)(n−2k−2)k(k-1){n\choose k}=n(n-1){n-2\choose k-2}

For n=20n=20,

A=20⋅19∑k=420(−1)k(18k−2)A=20\cdot19\sum_{k=4}^{20}(-1)^k {18\choose k-2}

Put j=k−2j=k-2. Then

A=20⋅19∑j=218(−1)j(18j)A=20\cdot19\sum_{j=2}^{18}(-1)^j {18\choose j}

Using

∑j=018(−1)j(18j)=(1−1)18=0\sum_{j=0}^{18}(-1)^j{18\choose j}=(1-1)^{18}=0

we get

∑j=218(−1)j(18j)=−(1−18)=17\sum_{j=2}^{18}(-1)^j {18\choose j}=-(1-18)=17

Hence,

A=20⋅19⋅17=6460A=20\cdot19\cdot17=6460

Now compute BB. Since

∑k=020(−1)kk(20k)=0\sum_{k=0}^{20}(-1)^k k{20\choose k}=0

therefore

B=−∑k=03(−1)kk(20k)B=-\sum_{k=0}^{3}(-1)^k k{20\choose k}

So,

B=−(0−20+380−3420)=3060B=-\big(0-20+380-3420\big)=3060

Now compute CC. Since

∑k=020(−1)k(20k)=0\sum_{k=0}^{20}(-1)^k {20\choose k}=0

therefore

C=−∑k=03(−1)k(20k)C=-\sum_{k=0}^{3}(-1)^k {20\choose k}

So,

C=−(1−20+190−1140)=969C=-(1-20+190-1140)=969

Substitute all values:

S=A−4B+6CS=A-4B+6C S=6460−4(3060)+6(969)S=6460-4(3060)+6(969) S=6460−12240+5814=34S=6460-12240+5814=34

Therefore, the sum of the series is 3434.

Binomial Identity Split

Given:

S=∑k=420(−1)k(k−2)(k−3)(20k)S=\sum_{k=4}^{20}(-1)^k (k-2)(k-3){20\choose k}

Find: The numerical value of the sum.

The shortcut is to rewrite the quadratic factor as

(k−2)(k−3)=k(k−1)−4k+6(k-2)(k-3)=k(k-1)-4k+6

This works because sums involving k(k−1)(20k)k(k-1){20\choose k}, k(20k)k{20\choose k} and (20k){20\choose k} can be handled directly using standard alternating binomial identities from (1−1)n=0(1-1)^n=0.

After splitting,

S=A−4B+6CS=A-4B+6C

where the three parts evaluate quickly from the full alternating sums after subtracting the missing initial terms:

A=6460,B=3060,C=969A=6460,\quad B=3060,\quad C=969

Thus,

S=6460−12240+5814=34S=6460-12240+5814=34

Hence, the required value is 3434.

Common mistakes

  • A common mistake is writing the general term with the wrong sign pattern. The signs alternate and are represented by (−1)k(-1)^k when the index starts at k=4k=4. If the sign convention is shifted incorrectly, every subsequent sum changes. Always test the first term after defining the index.

  • Another mistake is expanding (k−2)(k−3)(k-2)(k-3) incorrectly. It must be

    (k−2)(k−3)=k2−5k+6=k(k−1)−4k+6(k-2)(k-3)=k^2-5k+6=k(k-1)-4k+6

    If this decomposition is wrong, the split into three standard binomial sums fails. Expand carefully before separating the series.

  • Students often use ∑k=020(−1)k(20k)=0\sum_{k=0}^{20}(-1)^k{20\choose k}=0 correctly but forget that the given sum starts from k=4k=4, not from k=0k=0. The missing terms for k=0,1,2,3k=0,1,2,3 must be subtracted separately. Do not apply the full identity directly without adjusting the lower limit.

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