1 + ((1/3) + (1/32)) + ((1/33 ) + (1/34) + (1/35)) + .... sum of the terms in the nth bracket =

Question

1 + ((1/3) + (1/32)) + ((1/33 ) + (1/34) + (1/35)) + .... sum of the terms in the nth bracket = (A) ((3n - 1)3/2.4n - 1) (B) (3n - 1/2.3(n - 1) (n+2) / 2 ) (C) (3n + 1/3.7n - 1) (D) (3n- 1/2n + 1)

Tardigrade Admin · Accepted Answer

Correct answer is (b) (3n - 1/2.3(n - 1) (n+2) / 2 )

Q. $1 + (\frac{1}{3} + \frac{1}{3 ^{2}}) + (\frac{1}{3 ^{3}} + \frac{1}{3 ^{4}} + \frac{1}{3 ^{5}}) + ....$ sum of the terms in the nth bracket =

$\frac{( 3 ^{n} - 1 ) ^{3}}{2. 4 ^{n - 1}}$

$\frac{3 ^{n} - 1}{2. 3 ^{(n - 1) (n + 2) /2}}$

$\frac{3 ^{n} + 1}{3. 7 ^{n - 1}}$

$\frac{3 ^{n - 1}}{2 ^{n + 1}}$

Correct answer is (b) $\frac{3 ^{n} - 1}{2. 3 ^{(n - 1) (n + 2) /2}}$

Q. 1+(31​+321​)+(331​+341​+351​)+.... sum of the terms in the nth bracket =

2.4n−1(3n−1)3​

2.3(n−1)(n+2)/23n−1​

3.7n−13n+1​

2n+13n−1​