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Question Number 56095 by gunawan last updated on 10/Mar/19

The coefficient of x^5  in the expansion of  (1+x)^(21)  +(1+x)^(22) +...+(1+x)^(30)   is

$$\mathrm{The}\:\mathrm{coefficient}\:\mathrm{of}\:{x}^{\mathrm{5}} \:\mathrm{in}\:\mathrm{the}\:\mathrm{expansion}\:\mathrm{of} \\ $$$$\left(\mathrm{1}+{x}\right)^{\mathrm{21}} \:+\left(\mathrm{1}+{x}\right)^{\mathrm{22}} +...+\left(\mathrm{1}+{x}\right)^{\mathrm{30}} \:\:\mathrm{is} \\ $$

Answered by tanmay.chaudhury50@gmail.com last updated on 10/Mar/19

(1+x)^n   let (r+1)th term contains x^5   nc_r x^r →so x^r =x^5 →r=5  now (1+x)^(21)  →21c_5 x^5   (1+x)^(22) →22c_5 x^5   ...  ...  required answer is  21c_5 +22c_5 +23c_5 +...+30c_5

$$\left(\mathrm{1}+{x}\right)^{{n}} \\ $$$${let}\:\left({r}+\mathrm{1}\right){th}\:{term}\:{contains}\:{x}^{\mathrm{5}} \\ $$$${nc}_{{r}} {x}^{{r}} \rightarrow{so}\:{x}^{{r}} ={x}^{\mathrm{5}} \rightarrow{r}=\mathrm{5} \\ $$$${now}\:\left(\mathrm{1}+{x}\right)^{\mathrm{21}} \:\rightarrow\mathrm{21}{c}_{\mathrm{5}} {x}^{\mathrm{5}} \\ $$$$\left(\mathrm{1}+{x}\right)^{\mathrm{22}} \rightarrow\mathrm{22}{c}_{\mathrm{5}} {x}^{\mathrm{5}} \\ $$$$... \\ $$$$... \\ $$$${required}\:{answer}\:{is} \\ $$$$\mathrm{21}{c}_{\mathrm{5}} +\mathrm{22}{c}_{\mathrm{5}} +\mathrm{23}{c}_{\mathrm{5}} +...+\mathrm{30}{c}_{\mathrm{5}} \\ $$

Answered by mr W last updated on 10/Mar/19

(1+x)^(21)  +(1+x)^(22) +...+(1+x)^(30)   =(((1+x)^(21) [(1+x)^(10) −1])/((1+x)−1))  =(((1+x)^(31) −(1+x)^(21) )/x)  coef. of x^5  term is:  C_6 ^(31) −C_6 ^(21) =682017

$$\left(\mathrm{1}+{x}\right)^{\mathrm{21}} \:+\left(\mathrm{1}+{x}\right)^{\mathrm{22}} +...+\left(\mathrm{1}+{x}\right)^{\mathrm{30}} \\ $$$$=\frac{\left(\mathrm{1}+{x}\right)^{\mathrm{21}} \left[\left(\mathrm{1}+{x}\right)^{\mathrm{10}} −\mathrm{1}\right]}{\left(\mathrm{1}+{x}\right)−\mathrm{1}} \\ $$$$=\frac{\left(\mathrm{1}+{x}\right)^{\mathrm{31}} −\left(\mathrm{1}+{x}\right)^{\mathrm{21}} }{{x}} \\ $$$${coef}.\:{of}\:{x}^{\mathrm{5}} \:{term}\:{is}: \\ $$$${C}_{\mathrm{6}} ^{\mathrm{31}} −{C}_{\mathrm{6}} ^{\mathrm{21}} =\mathrm{682017} \\ $$

Commented by gunawan last updated on 10/Mar/19

wow thank you Sir

$$\mathrm{wow}\:\mathrm{thank}\:\mathrm{you}\:\mathrm{Sir} \\ $$

Commented by malwaan last updated on 10/Mar/19

what is the formula   that you used ? please

$$\boldsymbol{{what}}\:\boldsymbol{{is}}\:\boldsymbol{{the}}\:\boldsymbol{{formula}}\: \\ $$$$\boldsymbol{{that}}\:\boldsymbol{{you}}\:\boldsymbol{{used}}\:?\:\boldsymbol{{please}} \\ $$

Commented by mr W last updated on 10/Mar/19

S=(1+x)^(21)  +(1+x)^(22) +...+(1+x)^(30)   is a geometric progression with  a_1 =(1+x)^(21)   q=(1+x)  n=10  S=((a_1 (q^n −1))/(q−1))=(((1+x)^(21) [(1+x)^(10) −1])/((1+x)−1))  =(((1+x)^(31) −(1+x)^(21) )/x)  x^6  term of (1+x)^(31)  is C_6 ^(31) x^6   x^6  term of (1+x)^(21)  is C_6 ^(21) x^6   x^5  term of S is ((C_6 ^(31) x^6 −C_6 ^(21) x^6 )/x)=(C_6 ^(31) −C_6 ^(21) )x^5

$${S}=\left(\mathrm{1}+{x}\right)^{\mathrm{21}} \:+\left(\mathrm{1}+{x}\right)^{\mathrm{22}} +...+\left(\mathrm{1}+{x}\right)^{\mathrm{30}} \\ $$$${is}\:{a}\:{geometric}\:{progression}\:{with} \\ $$$${a}_{\mathrm{1}} =\left(\mathrm{1}+{x}\right)^{\mathrm{21}} \\ $$$${q}=\left(\mathrm{1}+{x}\right) \\ $$$${n}=\mathrm{10} \\ $$$${S}=\frac{{a}_{\mathrm{1}} \left({q}^{{n}} −\mathrm{1}\right)}{{q}−\mathrm{1}}=\frac{\left(\mathrm{1}+{x}\right)^{\mathrm{21}} \left[\left(\mathrm{1}+{x}\right)^{\mathrm{10}} −\mathrm{1}\right]}{\left(\mathrm{1}+{x}\right)−\mathrm{1}} \\ $$$$=\frac{\left(\mathrm{1}+{x}\right)^{\mathrm{31}} −\left(\mathrm{1}+{x}\right)^{\mathrm{21}} }{{x}} \\ $$$${x}^{\mathrm{6}} \:{term}\:{of}\:\left(\mathrm{1}+{x}\right)^{\mathrm{31}} \:{is}\:{C}_{\mathrm{6}} ^{\mathrm{31}} {x}^{\mathrm{6}} \\ $$$${x}^{\mathrm{6}} \:{term}\:{of}\:\left(\mathrm{1}+{x}\right)^{\mathrm{21}} \:{is}\:{C}_{\mathrm{6}} ^{\mathrm{21}} {x}^{\mathrm{6}} \\ $$$${x}^{\mathrm{5}} \:{term}\:{of}\:{S}\:{is}\:\frac{{C}_{\mathrm{6}} ^{\mathrm{31}} {x}^{\mathrm{6}} −{C}_{\mathrm{6}} ^{\mathrm{21}} {x}^{\mathrm{6}} }{{x}}=\left({C}_{\mathrm{6}} ^{\mathrm{31}} −{C}_{\mathrm{6}} ^{\mathrm{21}} \right){x}^{\mathrm{5}} \\ $$

Commented by malwaan last updated on 11/Mar/19

thank you sir

$$\mathrm{thank}\:\mathrm{you}\:\mathrm{sir} \\ $$

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