If tn-1-4n-2n-3 for n-1-2-3- then 1-t1-1-t2-1-t3-1-t2003-

The correct option is A 4006/3006 t_n=1/4(n+2)(n+3) #10; #10; ∑_n=1^n=20031/t_n=1/t_1+1/t_2+⋯+1/t_2003 #10; #10; ∑_n=1^n=20034/(n+2) ...

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Definition of Limit

If tn=1/4n+...

Question

If $t_{n} = \frac{1}{4} (n + 2) (n + 3)$ for $n = 1, 2, 3, . . .$ then $\frac{1}{t_{1}} + \frac{1}{t_{2}} + \frac{1}{t_{3}} + . . . \frac{1}{t_{2003}} =$

\frac{4006}{3006}

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\frac{4003}{3007}

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\frac{4006}{3008}

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\frac{4006}{3009}

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Solution

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Similar questions

t_{n} = \frac{1}{4} (n + 2) (n + 3)

n = 1, 2, 3....

\frac{1}{t_{1}} + \frac{1}{t_{2}} + \frac{1}{t_{3}} + . . . . . . . . . + \frac{1}{t_{2003}} =

t_{n} = \frac{1}{4} (n + 2) (n + 3)

n = 1, 2, 3

\frac{1}{t_{I}} + \frac{1}{t_{2}} + \frac{1}{t_{3}} + \dots \dots . . + \frac{1}{t_{20}} =

T_{1} = T_{2} = 1

T_{1}, T_{2}, T_{3}, . . ., T_{n}

T_{n + 2} = (T_{n + 1})^{- 1} + T_{n}

n = 1, 2, 3

T_{2019}

T_{1}, T_{2}, T_{3}, \dots

S_{1} = T_{1} + T_{2} + T_{3} + \dots + T_{n}

S_{2} = T_{2} + T_{4} + T_{6} + \dots + T_{n - 1}

n

\frac{S_{1}}{S_{2}}

(1^{2} - t_{1}) + (2^{2} - t_{2}) + . . . . . . . + (n^{2} - t_{n}) = \frac{1}{3} n (n^{2} - 1)

t_{n}

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