Question Number 207262 by hardmath last updated on 10/May/24
$$\mathrm{a}_{\boldsymbol{\mathrm{n}}} \:-\:\mathrm{number}\:\mathrm{series} \\ $$$$\mathrm{a}_{\mathrm{1}} \:=\:\mathrm{5} \\ $$$$\mathrm{d}\:=\:\mathrm{3} \\ $$$$\mathrm{a}_{\mathrm{2}} ^{\mathrm{2}} −\mathrm{a}_{\mathrm{1}} ^{\mathrm{2}} \:+\:\mathrm{a}_{\mathrm{4}} ^{\mathrm{2}} −\mathrm{a}_{\mathrm{3}} ^{\mathrm{2}} \:+\:\mathrm{a}_{\mathrm{6}} ^{\mathrm{2}} −\mathrm{a}_{\mathrm{5}} ^{\mathrm{2}} \:+\:...\:+\:\mathrm{a}_{\mathrm{10}} ^{\mathrm{2}} −\mathrm{a}_{\mathrm{9}} ^{\mathrm{2}} \:=\:? \\ $$
Answered by A5T last updated on 10/May/24
![(a_2 −a_1 )(a_2 +a_1 )=d×[2a_1 +d] a_(2n) ^2 −a_(2n−1) ^2 =(a_(2n) −a_(2n−1) )(a_(2n) +a_(2n−1) ) =d×[a_1 +(2n−1)d+a_1 +(2n−2)d] =d×[2a_1 +d(4n−3)] Σ_(n=1) ^5 a_(2n) ^2 −a_(2n−1) ^2 =d[10a_1 +d(1+5+9+13+17)] =3[50+3(45)]=555](Q207265.png)
$$\left({a}_{\mathrm{2}} −{a}_{\mathrm{1}} \right)\left({a}_{\mathrm{2}} +{a}_{\mathrm{1}} \right)={d}×\left[\mathrm{2}{a}_{\mathrm{1}} +{d}\right] \\ $$$${a}_{\mathrm{2}{n}} ^{\mathrm{2}} −{a}_{\mathrm{2}{n}−\mathrm{1}} ^{\mathrm{2}} =\left({a}_{\mathrm{2}{n}} −{a}_{\mathrm{2}{n}−\mathrm{1}} \right)\left({a}_{\mathrm{2}{n}} +{a}_{\mathrm{2}{n}−\mathrm{1}} \right) \\ $$$$={d}×\left[{a}_{\mathrm{1}} +\left(\mathrm{2}{n}−\mathrm{1}\right){d}+{a}_{\mathrm{1}} +\left(\mathrm{2}{n}−\mathrm{2}\right){d}\right] \\ $$$$={d}×\left[\mathrm{2}{a}_{\mathrm{1}} +{d}\left(\mathrm{4}{n}−\mathrm{3}\right)\right] \\ $$$$\underset{{n}=\mathrm{1}} {\overset{\mathrm{5}} {\sum}}{a}_{\mathrm{2}{n}} ^{\mathrm{2}} −{a}_{\mathrm{2}{n}−\mathrm{1}} ^{\mathrm{2}} ={d}\left[\mathrm{10}{a}_{\mathrm{1}} +{d}\left(\mathrm{1}+\mathrm{5}+\mathrm{9}+\mathrm{13}+\mathrm{17}\right)\right] \\ $$$$=\mathrm{3}\left[\mathrm{50}+\mathrm{3}\left(\mathrm{45}\right)\right]=\mathrm{555} \\ $$
Commented by hardmath last updated on 10/May/24
$$\mathrm{perfect}\:\mathrm{dear}\:\mathrm{professor}\:\mathrm{thankyou} \\ $$