Question and Answers Forum

All Questions      Topic List

None Questions

Previous in All Question      Next in All Question      

Previous in None      Next in None      

Question Number 30245 by .none. last updated on 19/Feb/18

1^2 −3^2 +5^2 −7^2 +•••+17^2 −19^2

$$\mathrm{1}^{\mathrm{2}} −\mathrm{3}^{\mathrm{2}} +\mathrm{5}^{\mathrm{2}} −\mathrm{7}^{\mathrm{2}} +\bullet\bullet\bullet+\mathrm{17}^{\mathrm{2}} −\mathrm{19}^{\mathrm{2}} \\ $$

Commented by Penguin last updated on 19/Feb/18

=Σ_(n=1) ^(10) (−1)^(n−1) (2n−1)^2

$$=\underset{{n}=\mathrm{1}} {\overset{\mathrm{10}} {\sum}}\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} \left(\mathrm{2}{n}−\mathrm{1}\right)^{\mathrm{2}} \\ $$

Commented by abdo imad last updated on 20/Feb/18

let put S_N (x)=Σ_(n=1) ^N (2n−1)^2 x^n and xfrom R −{1} S_N (x)=Σ_(n=1) ^N (4n^2 −4n +1)x^n =4Σ_(n=1) ^N n^2 x^n −4Σ_(n=) ^N n x^n + Σ_(n=1) ^N x^n =4A_n (x) −4B_n (x) +C_n (x) we have C_n (x)=x Σ_(n=1) ^N x^(n−1) =xΣ_(n=0) ^(N−1) x^n =x ((1−x^N )/(1−x)) ⇒((dC_n (x))/dx)= (d/dx)(((x−x^(N+1) )/(1−x)))=Σ_(n=1) ^N nx^(n−1) =((B_n (x))/x) ⇒ B_n (x)=x (d/dx)( ((x −x^(N+1) )/(1−x)))=x (((1−(N+1)x^N +(x−x^(N+1) ))/((1−x^ )^2 ))... also ((dB_n (x))/dx)=Σ_(n=1) ^N n^2 x^(n−1) =(1/x)Σ_(n=1) ^N n^2 x^n ⇒ A_n (x)=x ((dB_n (x))/dx) ...after finding A_n ,B_n ,C_n take x=−1 and N=20 ....

$${let}\:{put}\:{S}_{{N}} \left({x}\right)=\sum_{{n}=\mathrm{1}} ^{{N}} \left(\mathrm{2}{n}−\mathrm{1}\right)^{\mathrm{2}} {x}^{{n}} \:\:\:{and}\:{xfrom}\:{R}\:−\left\{\mathrm{1}\right\} \\ $$$${S}_{{N}} \left({x}\right)=\sum_{{n}=\mathrm{1}} ^{{N}} \left(\mathrm{4}{n}^{\mathrm{2}} \:−\mathrm{4}{n}\:+\mathrm{1}\right){x}^{{n}} \\ $$$$=\mathrm{4}\sum_{{n}=\mathrm{1}} ^{{N}} {n}^{\mathrm{2}} {x}^{{n}} \:−\mathrm{4}\sum_{{n}=} ^{{N}} \:{n}\:{x}^{{n}} \:+\:\sum_{{n}=\mathrm{1}} ^{{N}} \:{x}^{{n}} \\ $$$$=\mathrm{4}{A}_{{n}} \left({x}\right)\:−\mathrm{4}{B}_{{n}} \left({x}\right)\:+{C}_{{n}} \left({x}\right)\:{we}\:{have} \\ $$$${C}_{{n}} \left({x}\right)={x}\:\sum_{{n}=\mathrm{1}} ^{{N}} {x}^{{n}−\mathrm{1}} ={x}\sum_{{n}=\mathrm{0}} ^{{N}−\mathrm{1}} \:{x}^{{n}} ={x}\:\frac{\mathrm{1}−{x}^{{N}} }{\mathrm{1}−{x}} \\ $$$$\Rightarrow\frac{{dC}_{{n}} \left({x}\right)}{{dx}}=\:\frac{{d}}{{dx}}\left(\frac{{x}−{x}^{{N}+\mathrm{1}} }{\mathrm{1}−{x}}\right)=\sum_{{n}=\mathrm{1}} ^{{N}} {nx}^{{n}−\mathrm{1}} =\frac{{B}_{{n}} \left({x}\right)}{{x}}\:\Rightarrow \\ $$$${B}_{{n}} \left({x}\right)={x}\:\frac{{d}}{{dx}}\left(\:\frac{{x}\:−{x}^{{N}+\mathrm{1}} }{\mathrm{1}−{x}}\right)={x}\:\frac{\left(\mathrm{1}−\left({N}+\mathrm{1}\right){x}^{{N}} \:+\left({x}−{x}^{{N}+\mathrm{1}} \right)\right.}{\left(\mathrm{1}−{x}^{} \right)^{\mathrm{2}} }... \\ $$$${also}\:\frac{{dB}_{{n}} \left({x}\right)}{{dx}}=\sum_{{n}=\mathrm{1}} ^{{N}} {n}^{\mathrm{2}} \:{x}^{{n}−\mathrm{1}} =\frac{\mathrm{1}}{{x}}\sum_{{n}=\mathrm{1}} ^{{N}} {n}^{\mathrm{2}} {x}^{{n}} \:\Rightarrow \\ $$$${A}_{{n}} \left({x}\right)={x}\:\frac{{dB}_{{n}} \left({x}\right)}{{dx}}\:...{after}\:{finding}\:{A}_{{n}} \:,{B}_{{n}} ,{C}_{{n}} \:{take}\:{x}=−\mathrm{1} \\ $$$${and}\:{N}=\mathrm{20}\:.... \\ $$$$ \\ $$

Commented by abdo imad last updated on 20/Feb/18

N=10.

$${N}=\mathrm{10}. \\ $$

Answered by ajfour last updated on 20/Feb/18

=(1−3)(1+3)+(5−7)(5+7)+.. ...+(17−19)(17+19) =−2(1+3+5+7+.....+17+19) =−2×10^2 = −200 .

$$=\left(\mathrm{1}−\mathrm{3}\right)\left(\mathrm{1}+\mathrm{3}\right)+\left(\mathrm{5}−\mathrm{7}\right)\left(\mathrm{5}+\mathrm{7}\right)+.. \\ $$$$\:\:\:\:...+\left(\mathrm{17}−\mathrm{19}\right)\left(\mathrm{17}+\mathrm{19}\right) \\ $$$$=−\mathrm{2}\left(\mathrm{1}+\mathrm{3}+\mathrm{5}+\mathrm{7}+.....+\mathrm{17}+\mathrm{19}\right) \\ $$$$=−\mathrm{2}×\mathrm{10}^{\mathrm{2}} \:=\:−\mathrm{200}\:\:. \\ $$

Commented by abdo imad last updated on 20/Feb/18

really you have a good looking for this qustion thanks...

$${really}\:{you}\:{have}\:{a}\:{good}\:{looking}\:{for}\:{this}\:{qustion}\:{thanks}... \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com