The sum of the series , is equal to __________. [2025]

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Solution

Q.
The sum of the series $2 \times 1 \times {}^{20}C_{4} - 3 \times 2 \times {}^{20}C_{5} + 4 \times 3 \times {}^{20}C_{6} - 5 \times 4 \times {}^{20}C_{7} + . . . + 18 \times 17 \times {}^{20}C_{20}$ , is equal to __________. [2025]

Ans.
(34)

Let $2 \times 1 \times {}^{20}C_{4} - 3 \times 2 \times {}^{20}C_{5} + . . . + 18 \times 17 \times {}^{20}C_{20} = A$

Using binomial theorem,

${(1 + x)}^{20} = {}^{20}C_{0} + {}^{20}C_{1} x + {}^{20}C_{2} x^{2} + {}^{20}C_{3} x^{3} + . . . + {}^{20}C_{20} x^{20}$

$\frac{{(1 + x)}^{20}}{x^{2}} = \frac{1}{x^{2}} + \frac{20}{x} + {}^{20}C_{2} + {}^{20}C_{3} x + . . . + {}^{20}C_{20} x^{18}$

Differentiate with respect to x, we get

$\frac{2 x {(1 + x)}^{19} (9 x - 1)}{x^{4}} = \frac{- 2}{x^{3}} - \frac{20}{x^{2}} + {}^{20}C_{3} + 2 \cdot {}^{20}C_{4} x + . . . + 18 {}^{20}C_{20} x^{17}$

$\frac{2 {(1 + x)}^{19} (9 x - 1)}{x^{3}} = \frac{- 2}{x^{3}} - \frac{20}{x^{2}} + {}^{20}C_{3} + 2 \cdot {}^{20}C_{4} x + . . . + 18 {}^{20}C_{20} x^{17}$

Again, differentiate with respect to x, we get

$\frac{2 [x^{3} {(1 + x)}^{19} \cdot 9 + x^{3} (9 x - 1) \cdot 19 {(1 + x)}^{18} - {(1 + x)}^{19} (9 x - 1) \cdot 3 x^{2}]}{x^{6}}$

$= \frac{6}{x^{4}} + \frac{40}{x^{3}} + 2 \cdot 1 \cdot {}^{20}C_{4} + 3 \cdot 2 \cdot {}^{20}C_{5} x + . . . + 18 \cdot 17 \cdot {}^{20}C_{20} x^{16}$

Put x = –1 in above equation,

0 = 6 – 40 + A

$\Rightarrow$ A = 34.

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