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JEE Main · 2025 · 07 Apr 2025 · Shift 2 · Binomial Theorem

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

Answer: 34

Solution

Each term is

(r - 2) (r - 3) (r 20) (- 1)^{r}

for

r = 4

to

20

. Consider

g (x) = x^{- 2} (1 + x)^{20} = \sum_{r} (r 20) x^{r - 2}

, so that

g^{''} (x) = \sum_{r} (r - 2) (r - 3) (r 20) x^{r - 4}

and

g^{''} (- 1) = \sum_{r = 0}^{20} (r - 2) (r - 3) (r 20) (- 1)^{r}

. Since

g^{''} (x) = 6 x^{- 4} (1 + x)^{20} - 80 x^{- 3} (1 + x)^{19} + 380 x^{- 2} (1 + x)^{18}

, every term carries a positive power of

(1 + x)

, so

g^{''} (- 1) = 0

. The

r = 0

and

r = 1

terms contribute

6

and

- 40

, so the required sum is

g^{''} (- 1) - 6 + 40 = 34

.

JEE Main · 2025 · 07 Apr 2025 · Shift 2 · Integer answer

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