Question

33) Find the sum of infinite terms of
1+4/5+7/25+10/125+13/625.....​

Answer 1

Answer:

35/16.

Step-by-step explanation:

S= 1 + 4/5 + 7/25 + 10/125 + 13/625 + .....   ----------------------  [1]

Multiply by 1/5 on both sides.

S/5 = 1/5 + 4/25 + 7/125 + 10/625 + ........

=> S/5 = 0 + 1/5 + 4/25 + 7/125 + 10/625 + ........  --------------  [2].

Subtract [2] from [1] ....

S - S/5 = (1 - 0) + (4/5 - 1/5) + (7/25 - 4/25) + (10/125 - 7/125) + (13/625 -  

                                                                                                     10/625) +  ...

=> 4S/5 = 1 + 3/5 + 3/25 + 3/125 + 3/625 + ........

=> 4S/5 = 1 + 3/5 (1 + 1/5 + 1/25 + 1/125 + ....)

Consider 1 + 1/5 + 1/25 + /125 + ... is an infinite GP with first term a = 1 and common ratio = 1/5,  whose sum to infinite terms = a/(1 - r) = 1/(1 - 1/5)  = 5/4.

=> 4S/5 = 1 + 3/5 *5/4 = 7/4

=> S = 7/4 * 5/4 = 35/16.

So the sum of given sequence to Infinite series = 35/16.