Question

Find the 18th and 25th terms of sequence given by
Tn = { n(n+2) if n is even and 4n/n^2+1 if n is odd ​

Answer 1

Answer:

hope you got your answer

Answer 2

$Given \:i) T_{n} = n(n+2) \: if \: n \:is \: even$

$and \:ii) T_{n} = \frac{4n}{n^{2} + 1 } \: if \: n \:is \: odd$

$If \: n = 18 \: \pink { (even ) }$

$T_{18} = 18(18+2)$

$= 18 \times 20$

$\green { = 360 }$

$If \: n = 25 \: \blue { (odd ) }$

$T_{25} = \frac{4\times 25}{25^{2} + 1 }$

$= \frac{ 100}{625 + 1 }$

$= \frac{100}{626}$

$\green { = \frac{ 50}{313} }$

Therefore.,

$\red{ 18^{th} \:term \: in \: sequence }\green { = 360 }$

$\red{ 25^{th} \:term \: in \: sequence }\green { = \frac{50}{313} }$

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