Question

$\huge\red{\boxed{\bf{\mid{\overline{\underline{Test\:Your\:Ability :}}}}\mid}}$
Find the sum of the series $\sf 1 + {}^{4}\!/{}_{7} +{}^{9}\!/{}_{7^2} + {}^{16}\!/{}_{7^3} + ..... \infty$

$\sf A)\:{}^{49}\!/{}_{27}\\\sf B)\:{}^{36}\!/{}_{49}\\\sf C)\:{}^{49}\!/{}_{36}\\\sf D)\:{}^{27}\!/{}_{49}$
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Answer 1

Step-by-step explanation:

$<marquee>$

$Find the sum of the series \: \sf 1 + {}^{4}\!/{}_{7} +{}^{9}\!/{}_{7^2} + {}^{16}\!/{}_{7^3} + ..... \infty \\ </p><p></p><p>\geqslant ]\sf A)\:{}^{49}\!/{}_{27}\\\sf B)\:{}^{36}\!/{}_{49}\\\sf C)\:{}^{49}\!/{}_{36}\\\sf D)\:{}^{27}\!/{}_{49}$

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

$</p><p></p><p>\geqslant ]\sf A)\:{}^{49}\!/27$

Answer 2

Answer:

This is a typical GP series problem.

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

Step-by-step explanation: