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JEE Main · 2023 · 06 Apr 2023 · Shift 2 · Binomial Theorem

If

(20)^{19} + 2 (21) (20)^{18} + 3 (21)^{2} (20)^{17} + \dots + 20 (21)^{19} = k (20)^{19}

, then k is equal to

Answer: 400

Solution

Factor out

(20)^{19}

:

k = 1 + 2 (\frac{21}{20}) + 3 (\frac{21}{20})^{2} + \dots + 20 (\frac{21}{20})^{19}

. Let

r = \frac{21}{20}

. Multiply by r:

k r = r + 2 r^{2} + \dots + 19 r^{19} + 20 r^{20}

. Subtract:

k (1 - r) = 1 + r + r^{2} + \dots + r^{19} - 20 r^{20} = \frac{r ^{20} - 1}{r - 1} - 20 r^{20}

. With

r - 1 = \frac{1}{20}

, this gives

k (\frac{- 1}{20}) = 20 r^{20} - 20 - 20 r^{20} = - 20

, so

k = 400

.

JEE Main · 2023 · 06 Apr 2023 · Shift 2 · Integer answer

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