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JEE Main · 2019 · 10 Jan 2019 · Shift 1 · Binomial Theorem

If

\sum_{i = 1}^{20} (\frac{^{20} C _{i - 1}}{^{20} C _{i} + ^{20} C _{i - 1}})^{3} = \frac{k}{21}

, then k equals:

(A)400
(B)50
(C)200
(D)100✓ Correct

Answer: (D)

Solution

Using

(i 20) + (i - 1 20) = (i 21)

, each term becomes

\frac{( i - 1 20 )}{( i 21 )}

, and the identity

\frac{( r - 1 n )}{( r n + 1 )} = \frac{r}{n + 1}

turns it into

\frac{i}{21}

. The sum is then

i = 1 \sum 20 (\frac{i}{21})^{3} = \frac{1}{2 1 ^{3}} (\frac{20 \cdot 21}{2})^{2} = \frac{21 0 ^{2}}{2 1 ^{3}} = \frac{44100}{9261} = \frac{100}{21}

. Matching this to

\frac{k}{21}

gives

k = 100

.

JEE Main · 2019 · 10 Jan 2019 · Shift 1 · Single correct option

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