Question

pls solve this Q.

THe infinite sum 1 + 4/7 + 9/7^2 + 16/7^3 + 25/7^4 + -----equals.

A)27/14

B)21/13

C)49/27

D)256/147

pls give me the soln to solve this Q.

Answer 1

Let

S = 1+4/7+9/7^2+16/7^3+………………..∞ (1)

S/7 = 1/7+4/7^2+9/7^3+16/7^4+……………. ∞ (2)

(1)-(2)=6S/7 =1+3/7+5/7^2+7/7^3+……………∞ (3)

6S/7 =1+3/7+5/7^2+7/7^3+……………∞ (4)

6S/7^2= 1/7+3/7^2+5/7^3+…………….∞ (5)

(4)-(5)= 36S/49=1+2/7+2/7^2+2/7^3+…………………∞ (6)

Now, 36S/49 = 1+2/7(1+1/7+1/7^2+……………..∞)

= 1+2/7(1/(1–1/7))= 1+2/7*7/6= 4/3

Therefore S = 49/27

Answer 2

The infinite sum 1 + 4/7 + 9/7^2 + 16/7^3 + 25/7^4 + -----equals to C)49/27.    

 

Step-by-step explanation:

Given:

$s= 1 + \frac{4}{7} + \frac{9}{7^2} +\frac{16}{7^3}+...$    [1]

multiplying equation 1 by 7,

$7s = 7 +\frac{4}{1}+\frac{9}{7}+\frac{16}{7^2}+...$              [2]

Subtracting eq 1 from eq 2, we get,

$7s-s=11-1+\frac{9-4}{7}+\frac{16-9}{7^2}+\frac{25-16}{7^3} +...$

$6s=10+\frac{5}{7}+ \frac{7}{7^2} +\frac{9}{7^3}+...$           [3]

multiplying eq 3 by 7,

$42s = 70 + 5+\frac{7}{7}+ \frac{9}{7}+...$                           [4]

subtracting eq 3 from eq 4 we get,

$42s-6s=75-10+\frac{7-5}{7}+ \frac{9-7}{7^2}+...$

$36s=65+\frac{2}{7} +\frac{2}{7^2}+...$     [Here the term are forming GP]

$36s=65+\frac{2}{7}\times \frac{1}{(1-\frac{1}{7} )}$

$36s=65+\frac{2}{7} \times\frac{1}{\frac{7-1}{7} }$

$36s=65+\frac{2}{7}\times \frac{7}{6}$

$36s=65+\frac{1}{3}$

$36s=\frac{195+1}{3}$

$36s=\frac{196}{3}$

$s=\frac{196}{36\times3}$

$s=\frac{196}{108}$

$s=\frac{49}{27}$