Question

If a₁, a₂, a₃,……. are in A.P. and a_i> 0 for each i, then

$\sum_{i = 1}^{n}\frac{n}{a_{i + 1}^{2/3} + a_{i + 1}^{1/3}a_{i}^{1/3} + a_{i}^{2/3}}$is equal to –

• A. $\frac{n}{a_{n}^{2/3} + a_{n}^{1/3} + a_{1}^{2/3}}$
• B. $\frac{n + 1}{a_{n}^{2/3} + a_{n}^{1/3} + a_{1}^{2/3}}$
• C. $\frac{n - 1}{a_{n}^{2/3} + a_{n}^{1/3}.a_{1}^{1/3} + a_{1}^{2/3}}$
• D. None of these

Answer 1

Answer: (C)

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

Answer 2

Let d be common difference of A.P. ̃ d = a_i – a_i–1

Now $\frac{1}{a_{i + 1}^{2/3} + a_{i + 1}^{1/3}.a_{i}^{1/3} + a_{i}^{2/3}}$= $\frac{a_{i + 1}^{1/3} - a_{i}^{1/3}}{a_{i + 1} - a_{i}}$= $\frac { 1 } { \mathrm {~d} }$ $\left\lbrack a_{i + 1}^{1/3} - a_{i}^{1/3} \right\rbrack$

Thus$\sum_{i = 1}^{n}\frac{n}{a_{i + 1}^{2/3} + a_{i + 1}^{1/3}.a_{i}^{1/3} + a_{i}^{2/3}}$
= $\frac{1}{d}$ $\sum_{i = 1}^{n - 1}\left( a_{i + 1}^{1/3} - a_{i}^{1/3} \right)$

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

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

\ [C] is correct.