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The-value-of-2-C-0-2-2-2-C-1-2-3-3-C-2-2-4-4-C-3-2-11-11-C-10-is-




Question Number 56140 by gunawan last updated on 11/Mar/19
The value of     2 C_0 +(2^2 /2)C_1 +(2^3 /3)C_2 +(2^4 /4)C_3 +...+(2^(11) /(11))C_(10)   is
Thevalueof2C0+222C1+233C2+244C3++21111C10is
Answered by tanmay.chaudhury50@gmail.com last updated on 11/Mar/19
(1+x)^(10) =c_0 x^0 +c_1 x+c_2 x^2 +...+c_(10) x^(10)   ∫(1+x)^(10) dx=∫(c_0 x^0 +c_1 x^ +c_2 x^2 +..+c_(10) x^(10) )dx  (((1+x)^(11) )/(11))=c_0 ×(x^1 /1)+c_1 ×(x^2 /2)+c_2 ×(x^3 /3)+...+c_(10) ×(x^(11) /(11))+C  put x=0 both side to find constant value of C  (1/(11))=C  now put both side x=2   (((1+2)^(11) )/(11))=c_0 ×(2^1 /1)+c_1 ×(2^2 /2)+c_2 ×(2^3 /3)+...+c_(10) ×(2^(11) /(11))+(1/(11))  so required answer is  (3^(11) /(11))−(1/(11))=c_0 ×(2^1 /1)+c_1 ×(2^2 /2)+c_2 ×(2^3 /3)+...+c_(10) ×(2^(11) /(11))  ((3^(11) −1)/(11))←it is the answer
(1+x)10=c0x0+c1x+c2x2++c10x10(1+x)10dx=(c0x0+c1x+c2x2+..+c10x10)dx(1+x)1111=c0×x11+c1×x22+c2×x33++c10×x1111+Cputx=0bothsidetofindconstantvalueofC111=Cnowputbothsidex=2(1+2)1111=c0×211+c1×222+c2×233++c10×21111+111sorequiredansweris31111111=c0×211+c1×222+c2×233++c10×21111311111itistheanswer

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