The sum of the series 2\cdot ^{20}\! C_{0}+5\cdot ^{20}\! C_{1}+8\cdot ^{20}\! C_{2}+11\cdot ^{20}\! C_{3}+....+62\cdot ^{20}\! C_{20} is equal to :
 

  • Option 1)

    2^{23}

  • Option 2)

    2^{24}

  • Option 3)

    2^{25}

     

  • Option 4)

    2^{26}

 

Answers (1)

2\cdot ^{20}\! C_{0}+5\cdot ^{20}\! C_{1}+8\cdot ^{20}\! C_{2}+11\cdot ^{20}\! C_{3}+....+62\cdot ^{20}\! C_{20}

Let S= 2\cdot ^{20}\! C_{0}+5\cdot ^{20}\! C_{1}+....+62\cdot ^{20}\! C_{20}\; \; \; \; (1)

and S=62\cdot ^{20}\! C_{20}+59\cdot ^{20}\! C_{59}+.......+2\cdot ^{20}\! C_{0}\; \; \; \; \; (2)

S+S=(1)+(2)=2S\; \; \; \; \; \; as(^{n}C_{r}=^{n}C_{n-r})

2S=64\left ( ^{20}C_{o}+^{20}C_{1}+......+^{20}C_{20} \right )

=64\cdot 2^{20}

S=32\cdot 2^{20}=2^{5}\cdot 2^{20}=2^{25}


Option 1)

2^{23}

Option 2)

2^{24}

Option 3)

2^{25}

 

Option 4)

2^{26}

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