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Log Function

If k=∑_r=0^n1...

Question

If k=∑_{r=0}^n1/C_r, then ∑_{r=0}^nr/C_r is equal to (1) nk (2) nk/2 (3) (n-1) (4) nk/3

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S_{n} = n \sum r = 0 \frac{1}{^{n} C_{r}}

t_{n} = n \sum r = 0 \frac{r}{^{n} C_{r}}

\frac{t_{n}}{s_{n}} =

n \sum k = 1 k 2^{- k} = 2 [1 - 2^{- n} - n {.2}^{- (n + 1)}]

\frac{{^{n} C}_{r} + 4 {^{n} C}_{r + 1} + 6 {^{n} C}_{r + 2} + 4 {^{n} C}_{r + 3} + {^{n} C}_{r + 4}}{{^{n} C}_{r} + 3 {^{n} C}_{r + 1} + 3 {^{n} C}_{r + 2} + {^{n} C}_{r + 3}} = \frac{n + k}{r + k}

. Find the value of k

n \in N

\infty \sum k = 1 k (1 - \frac{1}{n})^{k - 1} =

(\begin{matrix} n k \end{matrix})

^{n} C_{k}

[\begin{matrix} n k \end{matrix}] = ⎧ ⎪ ⎨ ⎪ ⎩ \begin{matrix} (\begin{matrix} n k \end{matrix}), & if 0 \leq k \leq n 0, & otherwise. \end{matrix}

A_{k} = 9 \sum i = 0 (\begin{matrix} 9 i \end{matrix}) [\begin{matrix} 12 12 - k + i \end{matrix}] + 8 \sum i = 0 (\begin{matrix} 8 i \end{matrix}) [\begin{matrix} 13 13 - k + i \end{matrix}]

A_{4} - A_{3} = 190 p,

p

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