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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

$$\mathrm{The}\:\mathrm{value}\:\mathrm{of}\:\:\: \\ $$$$\mathrm{2}\:{C}_{\mathrm{0}} +\frac{\mathrm{2}^{\mathrm{2}} }{\mathrm{2}}{C}_{\mathrm{1}} +\frac{\mathrm{2}^{\mathrm{3}} }{\mathrm{3}}{C}_{\mathrm{2}} +\frac{\mathrm{2}^{\mathrm{4}} }{\mathrm{4}}{C}_{\mathrm{3}} +...+\frac{\mathrm{2}^{\mathrm{11}} }{\mathrm{11}}{C}_{\mathrm{10}} \:\:\mathrm{is} \\ $$

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

$$\left(\mathrm{1}+{x}\right)^{\mathrm{10}} ={c}_{\mathrm{0}} {x}^{\mathrm{0}} +{c}_{\mathrm{1}} {x}+{c}_{\mathrm{2}} {x}^{\mathrm{2}} +...+{c}_{\mathrm{10}} {x}^{\mathrm{10}} \\ $$$$\int\left(\mathrm{1}+{x}\right)^{\mathrm{10}} {dx}=\int\left({c}_{\mathrm{0}} {x}^{\mathrm{0}} +{c}_{\mathrm{1}} {x}^{} +{c}_{\mathrm{2}} {x}^{\mathrm{2}} +..+{c}_{\mathrm{10}} {x}^{\mathrm{10}} \right){dx} \\ $$$$\frac{\left(\mathrm{1}+{x}\right)^{\mathrm{11}} }{\mathrm{11}}={c}_{\mathrm{0}} ×\frac{{x}^{\mathrm{1}} }{\mathrm{1}}+{c}_{\mathrm{1}} ×\frac{{x}^{\mathrm{2}} }{\mathrm{2}}+{c}_{\mathrm{2}} ×\frac{{x}^{\mathrm{3}} }{\mathrm{3}}+...+{c}_{\mathrm{10}} ×\frac{{x}^{\mathrm{11}} }{\mathrm{11}}+{C} \\ $$$${put}\:{x}=\mathrm{0}\:{both}\:{side}\:{to}\:{find}\:{constant}\:{value}\:{of}\:{C} \\ $$$$\frac{\mathrm{1}}{\mathrm{11}}={C} \\ $$$${now}\:{put}\:{both}\:{side}\:{x}=\mathrm{2}\: \\ $$$$\frac{\left(\mathrm{1}+\mathrm{2}\right)^{\mathrm{11}} }{\mathrm{11}}={c}_{\mathrm{0}} ×\frac{\mathrm{2}^{\mathrm{1}} }{\mathrm{1}}+{c}_{\mathrm{1}} ×\frac{\mathrm{2}^{\mathrm{2}} }{\mathrm{2}}+{c}_{\mathrm{2}} ×\frac{\mathrm{2}^{\mathrm{3}} }{\mathrm{3}}+...+{c}_{\mathrm{10}} ×\frac{\mathrm{2}^{\mathrm{11}} }{\mathrm{11}}+\frac{\mathrm{1}}{\mathrm{11}} \\ $$$${so}\:{required}\:{answer}\:{is} \\ $$$$\frac{\mathrm{3}^{\mathrm{11}} }{\mathrm{11}}−\frac{\mathrm{1}}{\mathrm{11}}={c}_{\mathrm{0}} ×\frac{\mathrm{2}^{\mathrm{1}} }{\mathrm{1}}+{c}_{\mathrm{1}} ×\frac{\mathrm{2}^{\mathrm{2}} }{\mathrm{2}}+{c}_{\mathrm{2}} ×\frac{\mathrm{2}^{\mathrm{3}} }{\mathrm{3}}+...+{c}_{\mathrm{10}} ×\frac{\mathrm{2}^{\mathrm{11}} }{\mathrm{11}} \\ $$$$\frac{\mathrm{3}^{\mathrm{11}} −\mathrm{1}}{\mathrm{11}}\leftarrow{it}\:{is}\:{the}\:{answer} \\ $$

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