Sum of n terms of the series (2/5)+(22/25)+(122/125)+(622/625)+ ldots ldots .. is

Question

Sum of n terms of the series (2/5)+(22/25)+(122/125)+(622/625)+ ldots ldots .. is (A) 1-3 ⋅ 5-n (B) n -(3/4)+(3/4) 5- n  (C) n +(3/4)-(3/4) 5- n  (D) n-(3/4)+(1/4) 5-n

Tardigrade Admin · Accepted Answer

Given sum =(1-(3/5))+(1-(3/25))+(1-(3/125))+(1-(3/625))+ ldots ldots = n -((3/5)+(3/52)+(3/53)+ ldots . . . n text terms )= n -((3/5)(1-(1/5 n ))/1-(1)5)= n -(3/4)(1-5- n )= n -(3/4)+(3/4) 5- n

Q. Sum of $n$ terms of the series $\frac{2}{5} + \frac{22}{25} + \frac{122}{125} + \frac{622}{625} + \dots\dots .$ . is

$1 - 3 \cdot 5^{- n}$

$n - \frac{3}{4} + \frac{3}{4} 5^{- n}$

$n + \frac{3}{4} - \frac{3}{4} 5^{- n}$

$n - \frac{3}{4} + \frac{1}{4} 5^{- n}$

Q. Sum of n terms of the series 52​+2522​+125122​+625622​+…….. is

1−3⋅5−n

n−43​+43​5−n

n+43​−43​5−n

n−43​+41​5−n

Given sum =(1−53​)+(1−253​)+(1−1253​)+(1−6253​)+…… =n−(53​+523​+533​+…...n terms )=n−1−51​53​(1−5n1​)​=n−43​(1−5−n)=n−43​+43​5−n

Q. Sum of $n$ terms of the series $\frac{2}{5} + \frac{22}{25} + \frac{122}{125} + \frac{622}{625} + \dots\dots .$ . is

$1 - 3 \cdot 5^{- n}$

$n - \frac{3}{4} + \frac{3}{4} 5^{- n}$

$n + \frac{3}{4} - \frac{3}{4} 5^{- n}$

$n - \frac{3}{4} + \frac{1}{4} 5^{- n}$