^{20}C_1 + 3 ^{20}C_2 + 3 ^{20}C_3 + ^{20}C_4 | vedclass

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(D) The given expression is $S = ^{20}C_1 + 3 \cdot ^{20}C_2 + 3 \cdot ^{20}C_3 + ^{20}C_4$.We can rewrite the expression as $S = (^{20}C_1 + ^{20}C_2

Vedclass · Accepted Answer

$^{23}C_4$

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(D) The given expression is $S = ^{20}C_1 + 3 \cdot ^{20}C_2 + 3 \cdot ^{20}C_3 + ^{20}C_4$.We can rewrite the expression as $S = (^{20}C_1 + ^{20}C_2) + 2 \cdot (^{20}C_2 + ^{20}C_3) + (^{20}C_3 + ^{20}C_4)$.Using the Pascal's identity $^{n}C_{r} + ^{n}C_{r-1} = ^{n+1}C_{r}$, we get:$S = ^{21}C_2 + 2 \cdot ^{21}C_3 + ^{21}C_4$.Further, $S = (^{21}C_2 + ^{21}C_3) + (^{21}C_3 + ^{21}C_4)$.Applying the identity again, $S = ^{22}C_3 + ^{22}C_4$.Finally, $S = ^{23}C_4$.

$^{20}C_1 + 3 ^{20}C_2 + 3 ^{20}C_3 + ^{20}C_4$ is equal to-

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