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Maths Question 8 – JEE-MAIN 2026

If 26(233(12C2)+255(12C4)+277(12C6)+⋯+21313(12C12))=313−α, then α is equal to :

Recognize that the given sum resembles the result of integrating a binomial expansion term by term.

Step 1: Express the sum using binomial expansion and integration✦ Active

Let the given sum be P. We know that (1+x)12+(1−x)12=2(12C0+12C2x2+12C4x4+⋯+12C12x12). Let f(x)=12((1+x)12+(1−x)12). Integrating f(x) from 0 to 2 gives:

∫02f(x)dx=[12C0x+12C2x33+12C4x55+⋯+12C12x1313]02

Evaluating the definite integral at the limits:

∫02f(x)dx=(12C0(2)+12C2233+12C4255+⋯+12C1221313)−(0)

This expression is 2⋅12C0+P. Since 12C0=1, we have ∫02f(x)dx=2+P. Therefore, P=∫02f(x)dx−2.

Step 2: Evaluate the definite integral○ Expand

Now, we evaluate the integral I=∫02f(x)dx:

I=∫0212((1+x)12+(1−x)12)dx=12[(1+x)1313−(1−x)1313]02

Substitute the limits of integration:

I=126[((1+2)13−(1−2)13)−((1+0)13−(1−0)13)] I=126[(313−(−1)13)−(113−113)] I=126[(313−(−1))−(1−1)] I=126[(313+1)−0]=313+126
💡 Teacher's Secret Hint

Be careful with the signs when evaluating (1−x)13 at the limits, especially for x=0 and x=2.

Step 3: Calculate α○ Expand

Substitute the value of I back into the expression for P:

P=I−2=313+126−2

The given equation is 26P=313−α. Multiply P by 26:

26P=26(313+126−2) 26P=(313+1)−26×2 26P=313+1−52 26P=313−51

Comparing this with 313−α, we find that α=51.

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