Question

Sa se afle suma 1+11+111+1111+...+111...1111 (ultimul termen are 2015 cifre).
Multumesc pentru ajutor :)

Answer 1

cum stii ca numarul abc=a*100+b*10+c, avem:

1+11+111+....+111....111=
=1+(1*10+1) +(1*100+1*10+1)+...+(1*10^2015+1*10^2014+...+1*100+1*10+1)
=10^0*(1+1+...+1)*2015 +10^1*(1+1+...+1)*2014 +10^2*(1+1+1+....+1)*2013  +....+10^2014 *(1+1) +10^2015*1=

10^0* 2015+10^1 *2015 +10^2 *2013 +...+10^2014*2 +10^2015 =S
 (o notam cu S de la suma)

apoi facem 10*S
10*S =10^1 *2015+10^2*2015 +....+10^2015 *2 +10^2016

apoi scadem 10*S - S lasand primul termen din S la urma :
9*S=10^1 *(2015-2014)+10^2 *(2014-2013) +...+10^2015 *(2-1) +10^2016 -10^0 *2015 =>
=> 9*S=10^1 +10^2+....+10^2016 -2015 
luam separat :
si notam 10^1 +10^2+...+10^2016 =S2(de la suma 2)
=> S2=10^1+...+10^2016
=> 10*S2= 10^2+...+10^2016+10^2017 => 10*S2-S2= 10^2017-10^1
=> 9*S2=10^2017-10 =>S2=(10^2017-10)/9
revenim la S : aveam: 9*S=S2-2015 => 9*S=(10^2017-10)/9 -2015 =>
=> S=(10^2017-10 -2015*9)/81

am notat ^ ca putere
grea problema pt clasele primare :-??