Question

aratati ca :a) 2*(1+3+3²+.........+3^99)<3^100; b)3*(1+4+4²+ ......................+4^2013)<4^2014. VĂ ROG REPEDE.

Answer 1

Fie n si k nr. naturale

S=1+n+n^2+n^3+...+n^k

Inmultim cu n:

nxS=n+n^2+n^3+...+n^k+n(k+1)

nxS=(1+n+n^2+n^3+...+n^k)+n(k+1) -1

nxS=S+n(k+1) -1

nxS-S=n(k+1) -1

S(n-1)=n(k+1) -1

S=(n(k+1) -1)/(n-1)

a)

n=3 si k=99

1+3+3^2+...+3^99=(3^100 -1)/2

2*(1+3+3^2+...+3^99)=2*(3^100 -1)/2=

3^100 -1<3^100

b)

n=4, k=2013

1+4+4^2+...+4^2013=(4^2014 -1)/3

3*(1+4+4^2+...+4^2013)=3*(4^2014 -1)/3=

4^2014 -1<4^2014