The sum of the series 2 x 1 x 20C4 - 3 x 2 x 20C5 + 4 x 3 x 20C6 - 5 x 4 x 20C7 + ... + 18 x 17 x 20C20, is equal to _________.

The correct answer is Option 34. Full mathematical solution steps and hints are accessible in the interactive StemCET workspace.

Maths Q21 – JEE-MAIN 2025

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Home Maths Binomial TheoremQuestion #18

The sum of the series

2 \times 1 \times^{20} C_{4} - 3 \times 2 \times^{20} C_{5} + 4 \times 3 \times^{20} C_{6} - 5 \times 4 \times^{20} C_{7} + \dots + 18 \times 17 \times^{20} C_{20}

, is equal to _______.

Enter Calculated Numerical Value:

Clue 1 of 3: Identify the general term

Express the given series in summation notation by finding a pattern for the terms, including the alternating sign and the indices of the binomial coefficients.

🧠 Full Solution Path

Step 1: Express the series in summation form✦ Active

The given series is $S = 2 \times 1 \times^{20} C_{4} - 3 \times 2 \times^{20} C_{5} + 4 \times 3 \times^{20} C_{6} - 5 \times 4 \times^{20} C_{7} + \dots + 18 \times 17 \times^{20} C_{20}$ . The general term can be written as $T_{k} = (- 1)^{k - 1} (k + 1) k^{20} C_{k + 3}$ . The sum runs from $k = 1$ (for the first term $2 \times 1 \times^{20} C_{4}$ ) to $k = 17$ (for the last term $18 \times 17 \times^{20} C_{20}$ ). Let $j = k + 3$ . Then $k = j - 3$ . The sum becomes:

S = \sum_{j = 4}^{20} (- 1)^{j - 4} (j - 2) (j - 3)^{20} C_{j}

Since $(- 1)^{j - 4} = (- 1)^{j}$ , we have:

S = \sum_{j = 4}^{20} (- 1)^{j} (j - 2) (j - 3)^{20} C_{j}

Step 2: Decompose the general term and apply identities○ Expand

We use the algebraic identity $(j - 2) (j - 3) = j^{2} - 5 j + 6 = j (j - 1) - 4 j + 6$ . Substituting this into the sum, we get:

S = \sum_{j = 4}^{20} (- 1)^{j} [j (j - 1) - 4 j + 6]^{20} C_{j}

This can be split into three separate sums:

S = \sum_{j = 4}^{20} (- 1)^{j} j (j - 1)^{20} C_{j} - 4 \sum_{j = 4}^{20} (- 1)^{j} j^{20} C_{j} + 6 \sum_{j = 4}^{20} (- 1)^{j}^{20} C_{j}

We use the following identities for binomial coefficients:

\sum_{r = 0}^{n} (- 1)^{r}^{n} C_{r} = 0 (n \geq 1)

\sum_{r = 0}^{n} (- 1)^{r} r^{n} C_{r} = 0 (n \geq 1)

\sum_{r = 0}^{n} (- 1)^{r} r (r - 1)^{n} C_{r} = 0 (n \geq 2)

Also, we use $j (j - 1)^{n} C_{j} = n (n - 1)^{n - 2} C_{j - 2}$ and $j^{n} C_{j} = n^{n - 1} C_{j - 1}$ .

💡 Teacher's Secret Hint

Remember to adjust the summation limits when using the general identities. The terms for $j = 0, 1, 2, 3$ might need to be subtracted or added.

Step 3: Calculate each sum and find the total○ Expand

Let $S_{1} = \sum_{j = 4}^{20} (- 1)^{j} j (j - 1)^{20} C_{j}$ . Using $j (j - 1)^{20} C_{j} = 20 \times 19^{18} C_{j - 2}$ and letting $m = j - 2$ :

S_{1} = 20 \times 19 \sum_{m = 2}^{18} (- 1)^{m + 2}^{18} C_{m} = 380 \sum_{m = 2}^{18} (- 1)^{m}^{18} C_{m}

Since $\sum_{m = 0}^{18} (- 1)^{m}^{18} C_{m} = 0$ , we have $\sum_{m = 2}^{18} (- 1)^{m}^{18} C_{m} = 0 - ((- 1)^{0}^{18} C_{0} + (- 1)^{1}^{18} C_{1}) = 0 - (1 - 18) = 17$ . So, $S_{1} = 380 \times 17 = 6460$ .

Let $S_{2} = \sum_{j = 4}^{20} (- 1)^{j} j^{20} C_{j}$ . Using $j^{20} C_{j} = 20^{19} C_{j - 1}$ and letting $m = j - 1$ :

S_{2} = 20 \sum_{m = 3}^{19} (- 1)^{m + 1}^{19} C_{m} = - 20 \sum_{m = 3}^{19} (- 1)^{m}^{19} C_{m}

Since $\sum_{m = 0}^{19} (- 1)^{m}^{19} C_{m} = 0$ , we have $\sum_{m = 3}^{19} (- 1)^{m}^{19} C_{m} = 0 - ((- 1)^{0}^{19} C_{0} + (- 1)^{1}^{19} C_{1} + (- 1)^{2}^{19} C_{2}) = 0 - (1 - 19 + \frac{19 \times 18}{2}) = 0 - (1 - 19 + 171) = - 153$ . So, $S_{2} = - 20 \times (- 153) = 3060$ .

Let $S_{3} = \sum_{j = 4}^{20} (- 1)^{j}^{20} C_{j}$ . Since $\sum_{j = 0}^{20} (- 1)^{j}^{20} C_{j} = 0$ :

S_{3} = 0 - ((- 1)^{0}^{20} C_{0} + (- 1)^{1}^{20} C_{1} + (- 1)^{2}^{20} C_{2} + (- 1)^{3}^{20} C_{3})

$S_{3} = 0 - (1 - 20 + \frac{20 \times 19}{2} - \frac{20 \times 19 \times 18}{3 \times 2 \times 1})$

$S_{3} = 0 - (1 - 20 + 190 - 1140) = 0 - (- 969) = 969$

Finally, substitute these values back into the expression for $S$ :

S = S_{1} - 4 S_{2} + 6 S_{3} = 6460 - 4 (3060) + 6 (969)

S = 6460 - 12240 + 5814

S = 12274 - 12240 = 34

💡 Teacher's Secret Hint

Carefully calculate each binomial coefficient and ensure correct signs when subtracting the initial terms from the full sum.

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Maths Question 21 – JEE-MAIN 2025

Clue 1 of 3: Identify the general term

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