Jumlah deret tak hingga dari 5+1+1/5+1/25+...adalah...

Bab Baris dan Deret

a = 5
r = [tex] \frac{1}{5} [/tex]

S∞ = [tex] \frac{a}{1 - r} [/tex]
S∞ = [tex] \frac{5}{1 - \frac{1}{5} } [/tex]
S∞ = [tex] \frac{5}{\frac{4}{5} } [/tex]
S∞ = [tex]\frac{25}{4} [/tex]
S∞ = [tex]6\frac{1}{4} [/tex]

5 + 1 + 1/5 + 1/25 + ... = ?

a = 5
r = U2 / U1 = 1 /5

Dit. S∞

S∞ = a / ( 1 - r )
S∞ = 5 / ( 1 - 1/5)
S∞ = 5 / (4/5)
S∞ = 5 × 5/4
S∞ = 25/4 atau 6 1/4

Bisa juga ,
mis ,
5 + 1 + 1/5 + 1/25 + ... = n
Maka ,
1 + 1/5 + 1/25 + ... = n - 5

5 + 1 + 1/5 + 1/25 + ... = n
Ruas kiri dpt dijadikan :
5( 1 + 1/5 + 1/25 + ...) = n
5(n - 5) = n
5n - 25 = n
5n - n = 25
4n = 25
n = 25/4 atau 6 1/4