A sequence a n is defined as -an- cases - - dfrac 1n- text when n text is even lendcases -where n - N- Then the sum of first four terms isA- 7-12-35-12-32C- 1-4D- 0

The correct option is A 712a_n=1n,when n is odd 1n,when n is even ∴ a_1 =1, a_2= 12, a_3=13, nbsp;a_4= 14 Required sum = 1 12+13 14 =712 ...

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Sum of n Terms

A sequence a ...

Question

A sequence ${a_{n}}$ is defined as
$a_{n} = ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩ \begin{matrix} \frac{1}{n}, when n is odd - \frac{1}{n}, when n is even \end{matrix}$
where $n \in N$ . Then the sum of first four terms is

\frac{7}{12}

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\frac{5}{12}

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0

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\frac{3}{4}

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{a_{n}}

a_{n} = ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩ \begin{matrix} \frac{1}{n}, when n is odd - \frac{1}{n^{2}}, when n is even \end{matrix}

n \in N

{a_{n}}

a_{n} = ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩ \begin{matrix} \frac{1}{n}, when n is odd - \frac{1}{n}, when n is even \end{matrix}

n \in N

{a_{n}}

a_{n} = ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩ \begin{matrix} \frac{1}{n}, when n is odd - \frac{1}{n^{2}}, when n is even \end{matrix}

n \in N

\frac{1}{2}

f : {1, 2, 3, . . . . . . . . .}

\to {0, \pm 1, \pm 2, . . . . . . .}

f (n) = ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩ \begin{matrix} \frac{n}{2}, if n is even - (\frac{n - 1}{2}), if n is odd \end{matrix}

f^{- 1} (- 100)

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