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Binomial Theorem question

2025 · 7 Apr · Shift 2 · Q46
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  5. /2025 · 7 Apr · Shift 2 · Q46

Binomial Theorem question

2025 · 7 Apr · Shift 2 · Q46

JEE MainMathematicsBinomial TheoremNumerical+4 / −1
The sum of the series 2×1×20C4−3×2×20C5+4×3×20C6−5×4×20C7+⋯⋯+18×17×20C202 \times 1 \times{ }^{20} \mathrm{C}_4-3 \times 2 \times{ }^{20} \mathrm{C}_5+4 \times 3 \times{ }^{20} \mathrm{C}_6-5 \times 4 \times{ }^{20} \mathrm{C}_7+\cdots \cdots+18 \times 17 \times{ }^{20} \mathrm{C}_{20}2×1×20C4​−3×2×20C5​+4×3×20C6​−5×4×20C7​+⋯⋯+18×17×20C20​, is equal to ‾\underline{\hspace{2cm}}​.
Numerical answer
View written solutionFree

Correct answer: 34

  1. Write the general term

The given series is 2⋅1(204)−3⋅2(205)+4⋅3(206)−⋯+18⋅17(2020).2\cdot 1\binom{20}{4}-3\cdot 2\binom{20}{5}+4\cdot 3\binom{20}{6}-\cdots+18\cdot 17\binom{20}{20}.2⋅1(420​)−3⋅2(520​)+4⋅3(620​)−⋯+18⋅17(2020​).

For r=4,5,6,…,20r=4,5,6,\dots,20r=4,5,6,…,20, the general term is (−1)r(r−2)(r−3)(20r).(-1)^r (r-2)(r-3)\binom{20}{r}.(−1)r(r−2)(r−3)(r20​).

So S=∑r=420(−1)r(r−2)(r−3)(20r).S=\sum_{r=4}^{20} (-1)^r (r-2)(r-3)\binom{20}{r}.S=∑r=420​(−1)r(r−2)(r−3)(r20​).

Since (r−2)(r−3)=r2−5r+6,(r-2)(r-3)=r^2-5r+6,(r−2)(r−3)=r2−5r+6, we get S=∑r=420(−1)r(r2−5r+6)(20r).S=\sum_{r=4}^{20} (-1)^r (r^2-5r+6)\binom{20}{r}.S=∑r=420​(−1)r(r2−5r+6)(r20​).

  1. Extend the sum from r=0r=0r=0 to 202020

Let T=∑r=020(−1)r(r−2)(r−3)(20r).T=\sum_{r=0}^{20} (-1)^r (r-2)(r-3)\binom{20}{r}.T=∑r=020​(−1)r(r−2)(r−3)(r20​).

Then S=T−∑r=03(−1)r(r−2)(r−3)(20r).S=T-\sum_{r=0}^{3} (-1)^r (r-2)(r-3)\binom{20}{r}.S=T−∑r=03​(−1)r(r−2)(r−3)(r20​).

  1. Evaluate TTT using binomial identities

We use the standard results from (1+x)20=∑r=020(20r)xr.(1+x)^{20}=\sum_{r=0}^{20} \binom{20}{r}x^r.(1+x)20=∑r=020​(r20​)xr.

Putting x=−1x=-1x=−1 gives ∑r=020(−1)r(20r)=(1−1)20=0.\sum_{r=0}^{20} (-1)^r \binom{20}{r}=(1-1)^{20}=0.∑r=020​(−1)r(r20​)=(1−1)20=0.

Also, ∑r=020r(−1)r(20r)=0,\sum_{r=0}^{20} r(-1)^r\binom{20}{r}=0,∑r=020​r(−1)r(r20​)=0, and ∑r=020r(r−1)(−1)r(20r)=0,\sum_{r=0}^{20} r(r-1)(-1)^r\binom{20}{r}=0,∑r=020​r(r−1)(−1)r(r20​)=0, because these come from derivatives of (1+x)20(1+x)^{20}(1+x)20 evaluated at x=−1x=-1x=−1, and each contains a factor (1+x)k(1+x)^{k}(1+x)k with k>0k>0k>0.

Now r2=r(r−1)+r,r^2=r(r-1)+r,r2=r(r−1)+r, so ∑r=020r2(−1)r(20r)=0+0=0.\sum_{r=0}^{20} r^2(-1)^r\binom{20}{r}=0+0=0.∑r=020​r2(−1)r(r20​)=0+0=0.

Hence T=∑r=020(−1)r(r2−5r+6)(20r)=0−5⋅0+6⋅0=0.T=\sum_{r=0}^{20} (-1)^r (r^2-5r+6)\binom{20}{r}=0-5\cdot 0+6\cdot 0=0.T=∑r=020​(−1)r(r2−5r+6)(r20​)=0−5⋅0+6⋅0=0.

Therefore, S=−∑r=03(−1)r(r−2)(r−3)(20r).S=-\sum_{r=0}^{3} (-1)^r (r-2)(r-3)\binom{20}{r}.S=−∑r=03​(−1)r(r−2)(r−3)(r20​).

  1. Compute the missing first four terms

For r=0r=0r=0: (−1)0(−2)(−3)(200)=6.(-1)^0(-2)(-3)\binom{20}{0}=6.(−1)0(−2)(−3)(020​)=6.

For r=1r=1r=1: (−1)1(−1)(−2)(201)=−40.(-1)^1(-1)(-2)\binom{20}{1}=-40.(−1)1(−1)(−2)(120​)=−40.

For r=2r=2r=2: (−1)2(0)(−1)(202)=0.(-1)^2(0)(-1)\binom{20}{2}=0.(−1)2(0)(−1)(220​)=0.

For r=3r=3r=3: (−1)3(1)(0)(203)=0.(-1)^3(1)(0)\binom{20}{3}=0.(−1)3(1)(0)(320​)=0.

So ∑r=03(−1)r(r−2)(r−3)(20r)=6−40=−34.\sum_{r=0}^{3} (-1)^r (r-2)(r-3)\binom{20}{r}=6-40=-34.∑r=03​(−1)r(r−2)(r−3)(r20​)=6−40=−34.

Thus, S=−(−34)=34.S=-(-34)=34.S=−(−34)=34.

  1. Final answer

34\boxed{34}34​

The derived answer matches the stored correct answer.

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