Binomial Theorem JEE 2026: Weighted Sums by Integration

What is the weighted-coefficient question in Binomial Theorem JEE 2026?

For this binomial theorem JEE 2026 question, option A, 675, is correct, but the supplied working contains a factor-of-two slip. The supplied question is labelled JEE Main, January 2026, Slot 1. The question bank tags it hard and gives an expected solving time of 120 seconds, not a measured student average.

For each positive integer degree, define the coefficients and weighted sum:

C_{r} = (\binom{m}{r}), P_{m} = \sum_{r = 0}^{m} (\binom{m}{r}) \frac{(- 2)^{r}}{r + 1} .

Evaluate:

\sum_{n = 1}^{25} \frac{1}{P_{2 n}} .

The expansion degree uses a dummy index, kept separate from the outer summation index.

The second term is the negative coefficient given in the question:

r = 1 : \frac{- 2}{2} = - 1, C_{1} \frac{- 2}{2} = - C_{1} .

How does integration create the denominator?

Integrate the binomial expansion term by term. Integration increases each power by one and divides by that new exponent, creating the required denominator. Start with the binomial identity:

(1 + x)^{m} = \sum_{r = 0}^{m} (\binom{m}{r}) x^{r} .

The operation that matches the coefficient pattern is:

\int x^{r} d x = \frac{x^{r + 1}}{r + 1} + C .

Use definite integrals so that no integration constant remains. Since this is a finite polynomial sum, integrating each term separately is valid:

\int_{0}^{t} (1 + x)^{m} d x = \sum_{r = 0}^{m} (\binom{m}{r}) \int_{0}^{t} x^{r} d x .

Evaluate the left side:

{[\frac{(1 + x)^{m + 1}}{m + 1}]}_{0}^{t} = \frac{(1 + t)^{m + 1} - 1}{m + 1} .

The right side becomes:

\sum_{r = 0}^{m} (\binom{m}{r}) \frac{t^{r + 1}}{r + 1} .

Divide both sides by the upper-limit variable, provided it is nonzero. This restores the original power needed in the weighted sum:

\sum_{r = 0}^{m} (\binom{m}{r}) \frac{t^{r}}{r + 1} = \frac{(1 + t)^{m + 1} - 1}{(m + 1) t}, t \neq 0 .

Negative limits are allowed because these are polynomial integrals. There is no infinite-series convergence condition to check before making the negative substitution.

How do you substitute negative two and check the even index?

Substitute first, then simplify the odd power before taking any reciprocal. The even index makes the exponent in the numerator odd, so the numerator becomes two, not one.

Substitution into the identity gives:

t = - 2 : P_{m} = \frac{(- 1)^{m + 1} - 1}{- 2 (m + 1)} = \frac{1 - (- 1)^{m + 1}}{2 (m + 1)} .

Now use the even degree required by the question:

m = 2 n, (- 1)^{2 n + 1} = - 1 .

Keep the numerator visible until it cancels:

P_{2 n} = \frac{1 - (- 1)}{2 (2 n + 1)} = \frac{2}{2 (2 n + 1)} = \frac{1}{2 n + 1} .

The supplied worked text loses the numerator two when it writes:

P_{2 n} = \frac{1}{2 (2 n + 1)} (incorrect) .

Its integration method and final answer of 675 are correct; this intermediate line is not. Carrying the extra denominator factor through consistently would produce 1350, not 675.

Check directly at the smallest allowed degree:

m = 2 : P_{2} = (\binom{20}{-}) (\binom{21}{+}) \frac{4}{3} (\binom{22}{=}) 1 - 2 + \frac{4}{3} = \frac{1}{3} .

The corrected formula gives one-third; the erroneous line gives one-sixth. This direct check catches the factor-of-two slip before it enters the final sum.

How do you take reciprocals and finish the sum?

The reciprocals are consecutive odd integers from 3 to 51. There are 25 terms, not 26, because the outer index starts at one. The corrected expression is nonzero, so taking its reciprocal is valid:

\frac{1}{P_{2 n}} = 2 n + 1 .

Thus the required sum is:

\sum_{n = 1}^{25} \frac{1}{P_{2 n}} = 3 + 5 + 7 + \dots + 51 .

Apply the arithmetic-progression formula using those endpoints and the term count:

S = \frac{25 (3 + 51)}{2} = 25 \times 27 = 675 .

Cross-check by splitting the summand. Keep the constant contribution from every term:

S = 2 \sum_{n = 1}^{25} n + 25 = 2 \times \frac{25 \times 26}{2} + 25 = 650 + 25 = 675 .

Answer: option A, 675.

How does dropping the index offset produce option B?

Dropping the added one from each reciprocal produces option B, 650. This error can occur even after deriving the correct binomial identity, if the offset disappears during substitution.

The incorrect calculation is:

\frac{1}{P_{2 n}} \overset{wrong}{=} 2 n, \sum_{n = 1}^{25} 2 n = 25 \times 26 = 650 .

The offset comes from the integrated exponent and survives the even-index substitution: $m + 1 ⟶ 2 n + 1 .$

It is not optional notation. Omitting one from each of 25 terms removes exactly 25 from the correct total: $675 - 25 = 650 .$

Check the first reciprocal before summing. The direct calculation gave one-third, so its reciprocal must be 3. A first term of 2 exposes the mistake immediately.

Which three related questions test the same integration method?

These original related practice questions test substitution, parity and the final index offset. They are not additional verified PYQs. Apply the derived identity, then compare your substitution and arithmetic with the worked checks.

Question 1: Evaluate the positive weighted sum.

\sum_{r = 0}^{5} (\binom{5}{r}) \frac{1}{r + 1} .

Choose degree five and substitute one into the identity:

m = 5, t = 1 : \frac{(1 + 1)^{6} - 1}{6 \times 1} = \frac{2^{6} - 1}{6} = \frac{64 - 1}{6} = \frac{21}{2} .

Question 2: Evaluate the alternating weighted sum.

\sum_{r = 0}^{5} (\binom{5}{r}) \frac{(- 2)^{r}}{r + 1} .

Use the same degree but substitute negative two:

m = 5, t = - 2 : P_{5} = \frac{1 - (- 1)^{6}}{12} = \frac{1 - 1}{12} = 0 .

Every odd degree gives an even exponent in this numerator, making the weighted sum zero. Odd-indexed reciprocals are therefore undefined, which explains why the original question deliberately uses even indices.

Question 3: Using the original definition, evaluate the shorter reciprocal sum.

\sum_{k = 1}^{10} \frac{1}{P_{2 k}} .

Substitute the even degree into the corrected formula, then sum ten terms:

m = 2 k : P_{2 k} = \frac{1}{2 k + 1} .

\sum_{k = 1}^{10} \frac{1}{P_{2 k}} = 3 + 5 + \dots + 21 = \frac{10 (3 + 21)}{2} = 120 .

Redo these three without looking at the checks. Write the numerator after substitution explicitly, and verify the first reciprocal before using the progression formula.

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Frequently asked questions

How do I handle 1/(r+1) in a binomial coefficient sum?

Integrate the expansion of (1+x)^m term by term from 0 to t. Each x^r becomes t^(r+1)/(r+1); dividing by t restores the required power t^r. For t nonzero, the resulting weighted sum is [(1+t)^(m+1)-1]/[(m+1)t].

What is the answer to the weighted-sum Binomial Theorem JEE 2026 question?

The answer is option A, 675. For the weighted sum P_m = sum from r = 0 to m of binom(m,r)(-2)^r/(r+1), integration gives P_(2n) = 1/(2n+1). Its reciprocals for n = 1 to 25 are 3, 5, ..., 51, whose sum is 675.

Why do I get 650 instead of 675 in the reciprocal sum?

You get 650 if you replace each reciprocal 2n+1 with 2n. The missing +1 contributes once for each of the 25 terms, so the total falls short by 25. Check the first reciprocal: it must be 3, not 2.

What is the factor-of-two error in the supplied solution?

After substituting the even degree, the numerator is 1-(-1) = 2, which cancels the factor 2 in the denominator. The correct result is P_(2n) = 1/(2n+1), not 1/[2(2n+1)]. A direct check gives P_2 = 1-2+4/3 = 1/3, confirming the correction.