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JEE Main 20268 AprilEvening Shift
Exam
JEE Main
Year
2026
Date
8 April
Shift
Evening Shift
Subject
Mathematics
Chapter
Binomial Theorem

JEE Main 2026 · 8 April · Evening Shift · Mathematics · Binomial Theorem

If 26((2^3)/(3) 12 2 + (2^5)/(5) 12 4 + (2^7)/(7) 12 6 + + 2^ 13 13...

If 26(233(122)+255(124)+277(126)+…+21313(1212))=313−α26\left(\dfrac{2^3}{3}\binom{12}{2} + \dfrac{2^5}{5}\binom{12}{4} + \dfrac{2^7}{7}\binom{12}{6} + \ldots + \dfrac{2^{13}}{13}\binom{12}{12}\right) = 3^{13} - \alpha, then α\alpha is equal to:

Options

  1. A4545
  2. B4848
  3. C5151
  4. D5454

Correct answer

C. 5151

Step-by-step solution

Let S=233 12C2+255 12C4+277 12C6+…+21313 12C12S = \dfrac{2^3}{3} \ ^{12}C_{2} + \dfrac{2^5}{5} \ ^{12}C_{4} + \dfrac{2^7}{7} \ ^{12}C_{6} + \ldots + \dfrac{2^{13}}{13} \ ^{12}C_{12}

Using the property 1r+1 nCr=1n+1 n+1Cr+1\dfrac{1}{r+1} \ ^{n}C_{r} = \dfrac{1}{n+1} \ ^{n+1}C_{r+1}, we can rewrite the terms:

S=2313 13C3+2513 13C5+2713 13C7+…+21313 13C13S = \dfrac{2^3}{13} \ ^{13}C_{3} + \dfrac{2^5}{13} \ ^{13}C_{5} + \dfrac{2^7}{13} \ ^{13}C_{7} + \ldots + \dfrac{2^{13}}{13} \ ^{13}C_{13}

13S=23 13C3+25 13C5+27 13C7+…+213 13C1313S = 2^3 \ ^{13}C_{3} + 2^5 \ ^{13}C_{5} + 2^7 \ ^{13}C_{7} + \ldots + 2^{13} \ ^{13}C_{13}

We know the binomial expansions:

(1+x)13= 13C0+ 13C1x+ 13C2x2+ 13C3x3+…+ 13C13x13(1+x)^{13} = \ ^{13}C_{0} + \ ^{13}C_{1}x + \ ^{13}C_{2}x^2 + \ ^{13}C_{3}x^3 + \ldots + \ ^{13}C_{13}x^{13}

(1−x)13= 13C0− 13C1x+ 13C2x2− 13C3x3+…− 13C13x13(1-x)^{13} = \ ^{13}C_{0} - \ ^{13}C_{1}x + \ ^{13}C_{2}x^2 - \ ^{13}C_{3}x^3 + \ldots - \ ^{13}C_{13}x^{13}

Subtracting the second equation from the first gives:

(1+x)13−(1−x)13=2( 13C1x+ 13C3x3+ 13C5x5+…+ 13C13x13)(1+x)^{13} - (1-x)^{13} = 2 \left( \ ^{13}C_{1}x + \ ^{13}C_{3}x^3 + \ ^{13}C_{5}x^5 + \ldots + \ ^{13}C_{13}x^{13} \right)

Substituting x=2x = 2:

(1+2)13−(1−2)13=2( 13C1(2)+ 13C3(23)+ 13C5(25)+…+ 13C13(213))(1+2)^{13} - (1-2)^{13} = 2 \left( \ ^{13}C_{1}(2) + \ ^{13}C_{3}(2^3) + \ ^{13}C_{5}(2^5) + \ldots + \ ^{13}C_{13}(2^{13}) \right)

313−(−1)13=2(26+13S)3^{13} - (-1)^{13} = 2 \left( 26 + 13S \right)

313+1=52+26S3^{13} + 1 = 52 + 26S

26S=313+1−5226S = 3^{13} + 1 - 52

26S=313−5126S = 3^{13} - 51

Comparing this with the given expression 26S=313−α26S = 3^{13} - \alpha, we get:

α=51\alpha = 51

Answer: 5151
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