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Question Number 48447 by rahul 19 last updated on 24/Nov/18

Find the sum Σ_(k=0) ^(10) ^(20) C_k .

Findthesum∑10k=020Ck.

Commented by rahul 19 last updated on 24/Nov/18

i think 2^(nd) last line is wrong... 3 times^(20) C_(10) has been taken ..... plss also tell final correct answer....

ithink2ndlastlineiswrong...3times20C10hasbeentaken.....plssalsotellfinalcorrectanswer....

Commented by rahul 19 last updated on 24/Nov/18

Commented by mr W last updated on 24/Nov/18

S=C_0 ^(20) +C_1 ^(20) +C_2 ^(20) +C_3 ^(20) +...+C_(10) ^(20) C_0 ^(20) +C_1 ^(20) +C_2 ^(20) +C_3 ^(20) +...+C_(10) ^(20) +C_(11) ^(20) +...+C_(20) ^(20) =2^(20) [(C_0 ^(20) +C_(20) ^(20) )+(C_1 ^(20) +C_(19) ^(20) )+...+(C_(10) ^(20) +C_(10) ^(20) )]−C_(10) ^(20) =2^(20) [2C_0 ^(20) +2C_1 ^(20) +...+2C_(10) ^(20) ]−C_(10) ^(20) =2^(20) 2[C_0 ^(20) +C_1 ^(20) +...+C_(10) ^(20) ]−C_(10) ^(20) =2^(20) 2S−C_(10) ^(20) =2^(20) ⇒S=2^(19) +(C_(10) ^(20) /2)

S=C020+C120+C220+C320+...+C1020C020+C120+C220+C320+...+C1020+C1120+...+C2020=220[(C020+C2020)+(C120+C1920)+...+(C1020+C1020)]−C1020=220[2C020+2C120+...+2C1020]−C1020=2202[C020+C120+...+C1020]−C1020=2202S−C1020=220⇒S=219+C10202

Commented by rahul 19 last updated on 24/Nov/18

thanks sir ����

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