Binomial Theorem JEE 2024: Weighted Sum and Its Flaws

What is the intended answer to this Binomial Theorem JEE 2024 question?

The intended answer is 3660, but this binomial theorem JEE 2024 question has two defects: the supplied solution omits a multiplier after differentiation, and the natural-number condition does not guarantee uniqueness. The geometric-sum method works once the multiplier is restored.

The sum contains sixty powers of one plus the variable, with the kth power multiplied by k:

S (x) = \sum_{k = 1}^{60} k (1 + x)^{k}, x \neq 0 .

Express the required quantity in the stated form for natural numbers, then find their intended sum:

3600 S (60) = a b^{b} + b, a, b \in N .

$Find a + b .$ The supplied question bank tags this numerical-answer question Hard; no answer choices were supplied. Source note: the bank labels it 2024 but records 2023-01-24, Morning, meaning 24 January 2023, Morning. The year and shift attribution need verification.

How do you remove the weights and sum the geometric progression?

Remove the numerical weights to obtain a geometric progression, then differentiate its sum to recover them. There is no need to expand sixty binomials.

Define the unweighted sum:

T (x) = \sum_{k = 1}^{60} (1 + x)^{k} .

Its first term and common ratio are both one plus the variable, and it contains 60 terms. Substituting into the geometric-sum formula gives:

G_{m} = \frac{A (r^{m} - 1)}{r - 1}, A = 1 + x, r = 1 + x, m = 60 .

T (x) = \frac{(1 + x) [(1 + x)^{60} - 1]}{(1 + x) - 1} .

Multiply out only the numerator’s outer factor:

T (x) = \frac{(1 + x)^{61} - (1 + x)}{x} .

The given nonzero condition permits this quotient form. Differentiate this quotient rather than expanding sixty powers.

Why must you multiply by one plus the variable after differentiating?

Differentiation produces the required weights but lowers every exponent by one. Multiplying by one plus the variable restores those powers. The supplied solution omits this multiplier; restoring it corrects the error without changing its geometric-series method.

Term-by-term differentiation gives:

T^{'} (x) = \sum_{k = 1}^{60} k (1 + x)^{k - 1} .

The required identity is therefore:

S (x) = (1 + x) T^{'} (x) .

The derivative alone is not the required sum. To differentiate the quotient, name its numerator:

N (x) = (1 + x)^{61} - (1 + x),

N^{'} (x) = 61 (1 + x)^{60} - 1 .

Apply the quotient rule explicitly:

T^{'} (x) = \frac{x [61 (1 + x)^{60} - 1] - [(1 + x)^{61} - (1 + x)]}{x^{2}} .

Collect the power terms and the remaining terms separately:

61 x (1 + x)^{60} - (1 + x)^{61} = (60 x - 1) (1 + x)^{60},

$- x + (1 + x) = 1 .$ Hence:

T^{'} (x) = \frac{(60 x - 1) (1 + x)^{60} + 1}{x^{2}} .

Restore the lost power and clear the denominator:

x^{2} S (x) = (60 x - 1) (1 + x)^{61} + (1 + x) .

How does substituting 60 give the intended answer 3660?

Substitution gives the intended natural numbers 3599 and 61, whose sum is 3660. The external factor 3600 cancels the squared denominator at the required input. There is no reason to calculate the large power numerically.

The arithmetic is:

1 + 60 = 61, 60 \times 60 - 1 = 3599, 60^{2} = 3600 .

Substitute into the identity:

3600 S (60) = 3599 \times 61^{61} + 61 .

Match the displayed structure with the requested form:

a b^{b} + b = 3599 \times 61^{61} + 61 .

a = 3599, b = 61 .

a + b = 3599 + 61 = 3660 .

This is the intended answer derived from the method. The supplied worked solution does not print an explicit final number.

However, natural numbers alone do not make the representation unique. Another valid choice is:

b = 1, a = 3600 S (60) - 1 .

This satisfies the same equation: $a \cdot 1^{1} + 1 = 3600 S (60) .$

Thus, 3660 follows from the intended structural matching. It is not uniquely determined by the stated condition.

What incorrect result comes from treating the derivative as the weighted sum?

Omitting the restoring multiplier makes the evaluated result smaller by a factor of 61. This is an incorrect working route, not an official distractor: the question uses numerical-answer format and supplies no options.

Compare the differentiated term with the required term:

\frac{d}{d x} (1 + x)^{k} = k (1 + x)^{k - 1},

Required term: k (1 + x)^{k} .

Omitting the multiplier leads to the false statement:

3600 S (60) \overset{incorrect}{=} 3599 \times 61^{60} + 1 .

The correct expression is exactly 61 times that result:

3599 \times 61^{61} + 61 = 61 (3599 \times 61^{60} + 1) .

A one-term check catches the error without quotient-rule algebra:

T (x) = 1 + x, T^{'} (x) = 1, S (x) = 1 + x .

The derivative alone does not reproduce even this single weighted term. The mistake is failing to restore the exponent, not an arithmetic slip.

Which three related questions help you practise this method?

Use a general weighted geometric sum, a binomial sum weighted by its index, and a binomial sum weighted by two consecutive index factors. All three below are original related practice questions, not verified PYQs.

How do you evaluate the general weighted sum at an integer input?

Remove the weights, sum the progression, then differentiate and restore the lost power. This gives a formula that can be evaluated at the required integer without expanding the sum.

Question 1: For an integer satisfying the restriction below, find the requested quantity:

S_{n} (x) = \sum_{k = 1}^{n} k (1 + x)^{k}, n \geq 2 .

$Find n^{2} S_{n} (n) .$ Worked answer:

T_{n} (x) = \frac{(1 + x)^{n + 1} - (1 + x)}{x}, x \neq 0,

S_{n} (x) = (1 + x) T_{n}^{'} (x) .

x^{2} S_{n} (x) = (n x - 1) (1 + x)^{n + 1} + (1 + x) .

Substitution gives:

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

How do you evaluate a binomial sum weighted by its index?

Differentiate the binomial expansion once, then multiply by the variable to restore the powers. For the sum below, this gives 2916.

Question 2: Evaluate:

\sum_{k = 1}^{6} k (\binom{6}{k}) 2^{k} .

Worked answer:

(1 + t)^{6} = \sum_{k = 0}^{6} (\binom{6}{k}) t^{k},

t \frac{d}{d t} (1 + t)^{6} = \sum_{k = 1}^{6} k (\binom{6}{k}) t^{k} = 6 t (1 + t)^{5} .

t = 2 : 6 \cdot 2 \cdot 3^{5} = 2916 .

How do you evaluate a binomial sum with two consecutive index factors?

Differentiate twice, then multiply by the variable’s square to restore the two lost powers. For the sum below, this gives 480.

Question 3: Evaluate:

\sum_{k = 2}^{6} k (k - 1) (\binom{6}{k}) .

Worked answer:

t^{2} \frac{d^{2}}{d t^{2}} (1 + t)^{6} = \sum_{k = 2}^{6} k (k - 1) (\binom{6}{k}) t^{k} = 30 t^{2} (1 + t)^{4} .

t = 1 : 30 \cdot 1^{2} \cdot 2^{4} = 480 .

Before substituting, compare one differentiated term with the required term. Multiply by the variable after one derivative, or its square after two, to restore the powers.

Next step: photograph a doubt on JEEnius and photograph any question you are stuck on and get a step-by-step solution, with a free-body diagram when the question needs one (20 free).

For a worked example of the same idea, see Adsorption and Colloids JEE 2017: Why A and B Are Correct.

Frequently asked questions

What is the answer to this Binomial Theorem JEE 2024 question?

The intended answer is 3660. The corrected calculation gives 3600S(60) = 3599 × 61^61 + 61, so structural matching yields a = 3599 and b = 61. However, the stated natural-number condition does not make this answer unique.

Why do we multiply by 1+x after differentiating the geometric sum?

Differentiating (1+x)^k gives k(1+x)^(k-1), whereas the required term is k(1+x)^k. Multiplying by 1+x restores the lost power, so S(x) = (1+x)T'(x). At x = 60, omitting this multiplier makes the evaluated expression smaller by a factor of 61.

Why is a+b not uniquely determined in this question?

The equation 3600S(60) = ab^b + b also permits b = 1 and a = 3600S(60) - 1. Both are natural numbers and satisfy the same equation. Thus, 3660 follows from the intended structural matching, not from a unique solution under the stated condition.

Is this a verified JEE 2024 question?

The year and shift attribution need verification. The supplied question bank labels the question 2024 but records 24 January 2023, Morning. These conflicting details do not establish it as a verified JEE 2024 question.