Q33 / 75 · JEE Main 2026 (08 April Shift 2) · Mathematics
If $26\left(\dfrac{2^3}{3}\binom{12}{2} + \dfrac{2^5}{5}\binom{12}{4} + \dfrac{2^7}{7}\binom{12}{6} + \ldots + \dfrac{2^{13}}{13}\binom{12}{12}\right) = 3^{13} - \alpha$, then $\alpha$ is equal to:
- A$45$
- B$48$
- Correct answerC$51$
- D$54$
Solution
Let $S = \dfrac{2^3}{3} \ ^{12}C_{2} + \dfrac{2^5}{5} \ ^{12}C_{4} + \dfrac{2^7}{7} \ ^{12}C_{6} + \ldots + \dfrac{2^{13}}{13} \ ^{12}C_{12}$
Using the property $\dfrac{1}{r+1} \ ^{n}C_{r} = \dfrac{1}{n+1} \ ^{n+1}C_{r+1}$, we can rewrite the terms:
$S = \dfrac{2^3}{13} \ ^{13}C_{3} + \dfrac{2^5}{13} \ ^{13}C_{5} + \dfrac{2^7}{13} \ ^{13}C_{7} + \ldots + \dfrac{2^{13}}{13} \ ^{13}C_{13}$
$13S = 2^3 \ ^{13}C_{3} + 2^5 \ ^{13}C_{5} + 2^7 \ ^{13}C_{7} + \ldots + 2^{13} \ ^{13}C_{13}$
We know the binomial expansions:
$(1+x)^{13} = \ ^{13}C_{0} + \ ^{13}C_{1}x + \ ^{13}C_{2}x^2 + \ ^{13}C_{3}x^3 + \ldots + \ ^{13}C_{13}x^{13}$
$(1-x)^{13} = \ ^{13}C_{0} - \ ^{13}C_{1}x + \ ^{13}C_{2}x^2 - \ ^{13}C_{3}x^3 + \ldots - \ ^{13}C_{13}x^{13}$
Subtracting the second equation from the first gives:
$(1+x)^{13} - (1-x)^{13} = 2 \left( \ ^{13}C_{1}x + \ ^{13}C_{3}x^3 + \ ^{13}C_{5}x^5 + \ldots + \ ^{13}C_{13}x^{13} \right)$
Substituting $x = 2$:
$(1+2)^{13} - (1-2)^{13} = 2 \left( \ ^{13}C_{1}(2) + \ ^{13}C_{3}(2^3) + \ ^{13}C_{5}(2^5) + \ldots + \ ^{13}C_{13}(2^{13}) \right)$
$3^{13} - (-1)^{13} = 2 \left( 26 + 13S \right)$
$3^{13} + 1 = 52 + 26S$
$26S = 3^{13} + 1 - 52$
$26S = 3^{13} - 51$
Comparing this with the given expression $26S = 3^{13} - \alpha$, we get:
$\alpha = 51$
Answer: $51$
Using the property $\dfrac{1}{r+1} \ ^{n}C_{r} = \dfrac{1}{n+1} \ ^{n+1}C_{r+1}$, we can rewrite the terms:
$S = \dfrac{2^3}{13} \ ^{13}C_{3} + \dfrac{2^5}{13} \ ^{13}C_{5} + \dfrac{2^7}{13} \ ^{13}C_{7} + \ldots + \dfrac{2^{13}}{13} \ ^{13}C_{13}$
$13S = 2^3 \ ^{13}C_{3} + 2^5 \ ^{13}C_{5} + 2^7 \ ^{13}C_{7} + \ldots + 2^{13} \ ^{13}C_{13}$
We know the binomial expansions:
$(1+x)^{13} = \ ^{13}C_{0} + \ ^{13}C_{1}x + \ ^{13}C_{2}x^2 + \ ^{13}C_{3}x^3 + \ldots + \ ^{13}C_{13}x^{13}$
$(1-x)^{13} = \ ^{13}C_{0} - \ ^{13}C_{1}x + \ ^{13}C_{2}x^2 - \ ^{13}C_{3}x^3 + \ldots - \ ^{13}C_{13}x^{13}$
Subtracting the second equation from the first gives:
$(1+x)^{13} - (1-x)^{13} = 2 \left( \ ^{13}C_{1}x + \ ^{13}C_{3}x^3 + \ ^{13}C_{5}x^5 + \ldots + \ ^{13}C_{13}x^{13} \right)$
Substituting $x = 2$:
$(1+2)^{13} - (1-2)^{13} = 2 \left( \ ^{13}C_{1}(2) + \ ^{13}C_{3}(2^3) + \ ^{13}C_{5}(2^5) + \ldots + \ ^{13}C_{13}(2^{13}) \right)$
$3^{13} - (-1)^{13} = 2 \left( 26 + 13S \right)$
$3^{13} + 1 = 52 + 26S$
$26S = 3^{13} + 1 - 52$
$26S = 3^{13} - 51$
Comparing this with the given expression $26S = 3^{13} - \alpha$, we get:
$\alpha = 51$
Answer: $51$
Hence, the correct answer is option: 'C'.