Menu Close

The-sum-of-n-terms-of-the-series-1-2-2-2-3-2-4-2-5-2-6-2-is-




Question Number 99351 by 659083337 last updated on 20/Jun/20
The sum of n terms of the series   1^2  − 2^2  + 3^2  − 4^2  + 5^2  − 6^2  +.... is
$$\mathrm{The}\:\mathrm{sum}\:\mathrm{of}\:{n}\:\mathrm{terms}\:\mathrm{of}\:\mathrm{the}\:\mathrm{series}\: \\ $$$$\mathrm{1}^{\mathrm{2}} \:−\:\mathrm{2}^{\mathrm{2}} \:+\:\mathrm{3}^{\mathrm{2}} \:−\:\mathrm{4}^{\mathrm{2}} \:+\:\mathrm{5}^{\mathrm{2}} \:−\:\mathrm{6}^{\mathrm{2}} \:+….\:\mathrm{is} \\ $$
Commented by I want to learn more last updated on 20/Jun/20
1^2  + 3^2  + 5^2  + ... − 2^2   −  4^2   −  6^2   −  ...         1^2  + 3^2  + 5^2  + ... to n terms  − (2^2   +  4^2   +  6^2   +  ...  to  n terms)       =     1^2  + 3^2  + 5^2  + ... +  (2n  −  1)^2   − [2^2   +  4^2   +  6^2   +  ...  (2n)^2 ]    =           (n/3)(4n^2  −  1)  −  ((2n(n  +  1)(2n  +  1))/3)    =                       (n/3)[(4n^2  −  1)  −  2(n  +  1)(2n  +  1)]    =                             (n/3)[4n^2  −  1  −  (4n^2   +  6n  +  2)]    =                             (n/3)(4n^2  −  1  −  4n^2   −  6n  −  2)    =                                   (n/3)(−  6n  −  3) ,      ⇒     − ((3n)/3)(2n  +  1)  Therefore,  1^2   −   2^2   +   3^2    −  4^2    +  5^2   −  6^2   +  ....    =   −  (2n^2   +  n)
$$\mathrm{1}^{\mathrm{2}} \:+\:\mathrm{3}^{\mathrm{2}} \:+\:\mathrm{5}^{\mathrm{2}} \:+\:…\:−\:\mathrm{2}^{\mathrm{2}} \:\:−\:\:\mathrm{4}^{\mathrm{2}} \:\:−\:\:\mathrm{6}^{\mathrm{2}} \:\:−\:\:… \\ $$$$\:\:\:\:\:\:\:\mathrm{1}^{\mathrm{2}} \:+\:\mathrm{3}^{\mathrm{2}} \:+\:\mathrm{5}^{\mathrm{2}} \:+\:…\:\mathrm{to}\:\mathrm{n}\:\mathrm{terms}\:\:−\:\left(\mathrm{2}^{\mathrm{2}} \:\:+\:\:\mathrm{4}^{\mathrm{2}} \:\:+\:\:\mathrm{6}^{\mathrm{2}} \:\:+\:\:…\:\:\mathrm{to}\:\:\mathrm{n}\:\mathrm{terms}\right) \\ $$$$\:\:\:\:\:=\:\:\:\:\:\mathrm{1}^{\mathrm{2}} \:+\:\mathrm{3}^{\mathrm{2}} \:+\:\mathrm{5}^{\mathrm{2}} \:+\:…\:+\:\:\left(\mathrm{2n}\:\:−\:\:\mathrm{1}\right)^{\mathrm{2}} \:\:−\:\left[\mathrm{2}^{\mathrm{2}} \:\:+\:\:\mathrm{4}^{\mathrm{2}} \:\:+\:\:\mathrm{6}^{\mathrm{2}} \:\:+\:\:…\:\:\left(\mathrm{2n}\right)^{\mathrm{2}} \right] \\ $$$$\:\:=\:\:\:\:\:\:\:\:\:\:\:\frac{\mathrm{n}}{\mathrm{3}}\left(\mathrm{4n}^{\mathrm{2}} \:−\:\:\mathrm{1}\right)\:\:−\:\:\frac{\mathrm{2n}\left(\mathrm{n}\:\:+\:\:\mathrm{1}\right)\left(\mathrm{2n}\:\:+\:\:\mathrm{1}\right)}{\mathrm{3}} \\ $$$$\:\:=\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{\mathrm{n}}{\mathrm{3}}\left[\left(\mathrm{4n}^{\mathrm{2}} \:−\:\:\mathrm{1}\right)\:\:−\:\:\mathrm{2}\left(\mathrm{n}\:\:+\:\:\mathrm{1}\right)\left(\mathrm{2n}\:\:+\:\:\mathrm{1}\right)\right] \\ $$$$\:\:=\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{\mathrm{n}}{\mathrm{3}}\left[\mathrm{4n}^{\mathrm{2}} \:−\:\:\mathrm{1}\:\:−\:\:\left(\mathrm{4n}^{\mathrm{2}} \:\:+\:\:\mathrm{6n}\:\:+\:\:\mathrm{2}\right)\right] \\ $$$$\:\:=\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{\mathrm{n}}{\mathrm{3}}\left(\mathrm{4n}^{\mathrm{2}} \:−\:\:\mathrm{1}\:\:−\:\:\mathrm{4n}^{\mathrm{2}} \:\:−\:\:\mathrm{6n}\:\:−\:\:\mathrm{2}\right) \\ $$$$\:\:=\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{\mathrm{n}}{\mathrm{3}}\left(−\:\:\mathrm{6n}\:\:−\:\:\mathrm{3}\right)\:,\:\:\:\:\:\:\Rightarrow\:\:\:\:\:−\:\frac{\mathrm{3n}}{\mathrm{3}}\left(\mathrm{2n}\:\:+\:\:\mathrm{1}\right) \\ $$$$\mathrm{Therefore}, \\ $$$$\mathrm{1}^{\mathrm{2}} \:\:−\:\:\:\mathrm{2}^{\mathrm{2}} \:\:+\:\:\:\mathrm{3}^{\mathrm{2}} \:\:\:−\:\:\mathrm{4}^{\mathrm{2}} \:\:\:+\:\:\mathrm{5}^{\mathrm{2}} \:\:−\:\:\mathrm{6}^{\mathrm{2}} \:\:+\:\:….\:\:\:\:=\:\:\:−\:\:\left(\mathrm{2n}^{\mathrm{2}} \:\:+\:\:\mathrm{n}\right) \\ $$
Commented by PRITHWISH SEN 2 last updated on 20/Jun/20
If n is even then the series becomes  (1^2 −2^2 )+(3^2 −4^2 )+......+[(n−3)^2 −(n−2)^2 ]+                                        [(n−1)^2 −n^2 ]  = −{3+7+......(2n−5)+(2n−1)}  it is an A.P with commn diff. 4 and no of terms  = (n/2)  then  S_(n/2) = −(n/4){2.3+((n/2)−1)4}= − ((n(n+1))/2)  now when n is odd then the series  = 1^2 +(3^2 −2^2 )+.......+{(n−2)^2 −(n−3)^2 }+                                                     {n^2 −(n−1)^2 }  = 1+5+9+......+(2n−5)+(2n−1)  it is also an A.P where  d=4  and no. of term=((n+1)/2)  then  S_((n+1)/2) = ((n+1)/4){2.1+(((n+1)/2) −1)4} = ((n(n+1))/2)  i.e S_n = (−1)^(n+1) .((n(n+1))/2)  now when n=5  1^2 −2^2 +3^2 −4^2 +5^2  = (−1)^6 .((5×6)/2)=15  when n=6  1^2 −2^2 +3^2 −4^2 +5^2 −6^2 =(−1)^7 .((6×7)/2)=−21  please check
$$\mathrm{If}\:\boldsymbol{\mathrm{n}}\:\boldsymbol{\mathrm{is}}\:\boldsymbol{\mathrm{even}}\:\boldsymbol{\mathrm{then}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{series}}\:\boldsymbol{\mathrm{becomes}} \\ $$$$\left(\mathrm{1}^{\mathrm{2}} −\mathrm{2}^{\mathrm{2}} \right)+\left(\mathrm{3}^{\mathrm{2}} −\mathrm{4}^{\mathrm{2}} \right)+……+\left[\left(\boldsymbol{\mathrm{n}}−\mathrm{3}\right)^{\mathrm{2}} −\left(\boldsymbol{\mathrm{n}}−\mathrm{2}\right)^{\mathrm{2}} \right]+ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left[\left(\boldsymbol{\mathrm{n}}−\mathrm{1}\right)^{\mathrm{2}} −\boldsymbol{\mathrm{n}}^{\mathrm{2}} \right] \\ $$$$=\:−\left\{\mathrm{3}+\mathrm{7}+……\left(\mathrm{2}\boldsymbol{\mathrm{n}}−\mathrm{5}\right)+\left(\mathrm{2}\boldsymbol{\mathrm{n}}−\mathrm{1}\right)\right\} \\ $$$$\boldsymbol{\mathrm{it}}\:\boldsymbol{\mathrm{is}}\:\boldsymbol{\mathrm{an}}\:\boldsymbol{\mathrm{A}}.\boldsymbol{\mathrm{P}}\:\boldsymbol{\mathrm{with}}\:\boldsymbol{\mathrm{commn}}\:\boldsymbol{\mathrm{diff}}.\:\mathrm{4}\:\boldsymbol{\mathrm{and}}\:\boldsymbol{\mathrm{no}}\:\boldsymbol{\mathrm{of}}\:\boldsymbol{\mathrm{terms}} \\ $$$$=\:\frac{\boldsymbol{\mathrm{n}}}{\mathrm{2}} \\ $$$$\boldsymbol{\mathrm{then}} \\ $$$$\boldsymbol{\mathrm{S}}_{\frac{\boldsymbol{\mathrm{n}}}{\mathrm{2}}} =\:−\frac{\boldsymbol{\mathrm{n}}}{\mathrm{4}}\left\{\mathrm{2}.\mathrm{3}+\left(\frac{\boldsymbol{\mathrm{n}}}{\mathrm{2}}−\mathrm{1}\right)\mathrm{4}\right\}=\:−\:\frac{\boldsymbol{\mathrm{n}}\left(\boldsymbol{\mathrm{n}}+\mathrm{1}\right)}{\mathrm{2}} \\ $$$$\boldsymbol{\mathrm{now}}\:\boldsymbol{\mathrm{when}}\:\boldsymbol{\mathrm{n}}\:\boldsymbol{\mathrm{is}}\:\boldsymbol{\mathrm{odd}}\:\boldsymbol{\mathrm{then}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{series}} \\ $$$$=\:\mathrm{1}^{\mathrm{2}} +\left(\mathrm{3}^{\mathrm{2}} −\mathrm{2}^{\mathrm{2}} \right)+…….+\left\{\left(\mathrm{n}−\mathrm{2}\right)^{\mathrm{2}} −\left(\mathrm{n}−\mathrm{3}\right)^{\mathrm{2}} \right\}+ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left\{\mathrm{n}^{\mathrm{2}} −\left(\mathrm{n}−\mathrm{1}\right)^{\mathrm{2}} \right\} \\ $$$$=\:\mathrm{1}+\mathrm{5}+\mathrm{9}+……+\left(\mathrm{2n}−\mathrm{5}\right)+\left(\mathrm{2n}−\mathrm{1}\right) \\ $$$$\boldsymbol{\mathrm{it}}\:\boldsymbol{\mathrm{is}}\:\boldsymbol{\mathrm{also}}\:\boldsymbol{\mathrm{an}}\:\boldsymbol{\mathrm{A}}.\boldsymbol{\mathrm{P}}\:\boldsymbol{\mathrm{where}} \\ $$$$\boldsymbol{\mathrm{d}}=\mathrm{4}\:\:\boldsymbol{\mathrm{and}}\:\boldsymbol{\mathrm{no}}.\:\boldsymbol{\mathrm{of}}\:\boldsymbol{\mathrm{term}}=\frac{\boldsymbol{\mathrm{n}}+\mathrm{1}}{\mathrm{2}} \\ $$$$\boldsymbol{\mathrm{then}} \\ $$$$\boldsymbol{\mathrm{S}}_{\frac{\boldsymbol{\mathrm{n}}+\mathrm{1}}{\mathrm{2}}} =\:\frac{\boldsymbol{\mathrm{n}}+\mathrm{1}}{\mathrm{4}}\left\{\mathrm{2}.\mathrm{1}+\left(\frac{\boldsymbol{\mathrm{n}}+\mathrm{1}}{\mathrm{2}}\:−\mathrm{1}\right)\mathrm{4}\right\}\:=\:\frac{\boldsymbol{\mathrm{n}}\left(\boldsymbol{\mathrm{n}}+\mathrm{1}\right)}{\mathrm{2}} \\ $$$$\boldsymbol{\mathrm{i}}.\boldsymbol{\mathrm{e}}\:\boldsymbol{\mathrm{S}}_{\boldsymbol{\mathrm{n}}} =\:\left(−\mathrm{1}\right)^{\boldsymbol{\mathrm{n}}+\mathrm{1}} .\frac{\boldsymbol{\mathrm{n}}\left(\boldsymbol{\mathrm{n}}+\mathrm{1}\right)}{\mathrm{2}} \\ $$$$\boldsymbol{\mathrm{now}}\:\boldsymbol{\mathrm{when}}\:\boldsymbol{\mathrm{n}}=\mathrm{5} \\ $$$$\mathrm{1}^{\mathrm{2}} −\mathrm{2}^{\mathrm{2}} +\mathrm{3}^{\mathrm{2}} −\mathrm{4}^{\mathrm{2}} +\mathrm{5}^{\mathrm{2}} \:=\:\left(−\mathrm{1}\right)^{\mathrm{6}} .\frac{\mathrm{5}×\mathrm{6}}{\mathrm{2}}=\mathrm{15} \\ $$$$\boldsymbol{\mathrm{when}}\:\boldsymbol{\mathrm{n}}=\mathrm{6} \\ $$$$\mathrm{1}^{\mathrm{2}} −\mathrm{2}^{\mathrm{2}} +\mathrm{3}^{\mathrm{2}} −\mathrm{4}^{\mathrm{2}} +\mathrm{5}^{\mathrm{2}} −\mathrm{6}^{\mathrm{2}} =\left(−\mathrm{1}\right)^{\mathrm{7}} .\frac{\mathrm{6}×\mathrm{7}}{\mathrm{2}}=−\mathrm{21} \\ $$$$\boldsymbol{\mathrm{please}}\:\boldsymbol{\mathrm{check}} \\ $$
Commented by Dwaipayan Shikari last updated on 20/Jun/20
Yes sir it is right
Commented by I want to learn more last updated on 20/Jun/20
sir, please use your  nth term to show few sum.  I want to understand it.
$$\mathrm{sir},\:\mathrm{please}\:\mathrm{use}\:\mathrm{your}\:\:\mathrm{nth}\:\mathrm{term}\:\mathrm{to}\:\mathrm{show}\:\mathrm{few}\:\mathrm{sum}. \\ $$$$\mathrm{I}\:\mathrm{want}\:\mathrm{to}\:\mathrm{understand}\:\mathrm{it}. \\ $$
Commented by I want to learn more last updated on 20/Jun/20
Thanks sir.
$$\mathrm{Thanks}\:\mathrm{sir}. \\ $$

Leave a Reply

Your email address will not be published. Required fields are marked *