Statement 1: k = 1 6 k ( 3 k - 2 k ) ( 3 k + 1 - 2 k + 1 ) = 2 Statement 2: k = 1 n ( k 3 - ( k - 1 ) 3 ) = n 3 for any natural number n

Question

Statement 1: $\sum_{k = 1}^{\infty} \frac{6^{k}}{(3^{k} - 2^{k}) (3^{k + 1} - 2^{k + 1})} = 2$

Statement 2: $\sum_{k = 1}^{n} (k^{3} - {(k - 1)}^{3}) = n^{3}$ for any natural number $n$

Smart Achievers · Accepted Answer

(1)

$\sum_{k = 1}^{\infty} \frac{6^{k}}{(3^{k} - 2^{k}) (3^{k + 1} - 2^{k + 1})} = \frac{1}{3} \sum_{k = 1}^{\infty} \frac{{(\frac{2}{3})}^{k}}{(1 - {(\frac{2}{3})}^{k}) (1 - {(\frac{2}{3})}^{k + 1})} = 2$

Solution

Q.
Statement 1: $\sum_{k = 1}^{\infty} \frac{6^{k}}{(3^{k} - 2^{k}) (3^{k + 1} - 2^{k + 1})} = 2$

Statement 2: $\sum_{k = 1}^{n} (k^{3} - {(k - 1)}^{3}) = n^{3}$ for any natural number $n$

1 Both statement 1 and statement 2 are true

2 Both statement 1 and statement 2 are false

3 Statement 1 is true, Statement 2 is false

4 Statement 1 is false, Statement 2 is true

Ans.
(1)

$\sum_{k = 1}^{\infty} \frac{6^{k}}{(3^{k} - 2^{k}) (3^{k + 1} - 2^{k + 1})} = \frac{1}{3} \sum_{k = 1}^{\infty} \frac{{(\frac{2}{3})}^{k}}{(1 - {(\frac{2}{3})}^{k}) (1 - {(\frac{2}{3})}^{k + 1})} = 2$

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