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If $\sum_{i = 1}^{n} {(\frac{{}^{n}C_{i - 1}}{{}^{n}C_{i} + {}^{n}C_{i - 1}})}^{3} = \frac{36}{13}$ then $n$ is equal to

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Important Questions on Binomial Theorem

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Mathematics>Algebra>Binomial Theorem>Properties of Binomial Coefficients

The number of ordered pairs

(r, k)

for which

6 \cdot {}^{35}{C_{r}} = (k^{2} - 3) \cdot {}^{36}{C_{r + 1}},

where

k

is an integer is

HARD

Mathematics>Algebra>Binomial Theorem>Properties of Binomial Coefficients

{\sum_{i = 1}^{20} (\frac{^{20} C_{i - 1}}{^{20} C_{i} +^{20} C_{i - 1}})}^{3} = \frac{k}{21},

then

k

equals

HARD

Mathematics>Algebra>Binomial Theorem>Properties of Binomial Coefficients

Suppose

\det [\begin{matrix} \sum_{k = 0}^{n} k & \sum_{k = 0}^{n} {}^{n}{C_{k}} k^{2} \\ \sum_{k = 0}^{n} {}^{n}{C_{k}} k & \sum_{k = 0}^{n} {}^{n}{C_{k}} 3^{k} \end{matrix}] = 0,

holds for some positive integer

n .

Then

\sum_{k = 0}^{n} \frac{{}^{n}{C_{k}}}{k + 1}

equals

EASY

Mathematics>Algebra>Binomial Theorem>Properties of Binomial Coefficients

c_{0}, c_{1}, c_{2}, \dots, ., c_{15}

are the binomial coefficients in the expansion of

(1 + x)^{15},

then the value of

\frac{c_{1}}{c_{0}} + 2 \frac{c_{2}}{c_{1}} + 3 \frac{c_{3}}{c_{2}} + \dots + 15 \frac{c_{15}}{c_{14}}

EASY

Mathematics>Algebra>Binomial Theorem>Properties of Binomial Coefficients

The value of

{}^{16}C_{9} + {}^{16}C_{10} - {}^{16}C_{6} - {}^{16}C_{7}

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Mathematics>Algebra>Binomial Theorem>Properties of Binomial Coefficients

The value of

\sum_{r = 1}^{10} r \cdot \frac{{}^{n}C_{r}}{{}^{n}C_{r - 1}}

is equal to

HARD

Mathematics>Algebra>Binomial Theorem>Properties of Binomial Coefficients

Sum of last

30

coefficents in the binomial expansion of

{(1 + x)}^{59}

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Mathematics>Algebra>Binomial Theorem>Properties of Binomial Coefficients

{}^{34}{C_{5}} + \sum_{r = 0}^{4} {}^{(38 - r)}{C_{4}} =

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Mathematics>Algebra>Binomial Theorem>Properties of Binomial Coefficients

The number of selection of

n

objects from

2 n

objects of which

n

are identical and the rest are different, is

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Mathematics>Algebra>Binomial Theorem>Properties of Binomial Coefficients

In order to get through in an examination of nine papers, a candidate has to pass in more papers than the number of papers in which he fails. The number of ways in which he can fail, in this examination is

EASY

Mathematics>Algebra>Binomial Theorem>Properties of Binomial Coefficients

(2 \leq r \leq n),

then

{}^{n}C_{r} + 2 \cdot {}^{n}C_{r + 1} + {}^{n}C_{r + 2}

is equal to

EASY

Mathematics>Algebra>Binomial Theorem>Properties of Binomial Coefficients

Total number of

6

-digit numbers in which only and all the five digits

1, 3, 5, 7

and

9

appears, is

EASY

Mathematics>Algebra>Binomial Theorem>Properties of Binomial Coefficients

For

2 \leq r \leq n, {}^{n}C_{r} + 2 {}^{n}C_{r - 1} + {}^{n}C_{r - 2}

is equal to

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Mathematics>Algebra>Binomial Theorem>Properties of Binomial Coefficients

The value of

\sum_{r = 0}^{20} {}^{50 - r}C_{6}

is equal to:

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Mathematics>Algebra>Binomial Theorem>Properties of Binomial Coefficients

{}^{n}C_{r - 1} = 10, {}^{n}C_{r} = 45

and

{}^{n}C_{r + 1} = 120,

then

r

equals

HARD

Mathematics>Algebra>Binomial Theorem>Properties of Binomial Coefficients

If in a regular polygon the number of diagonals is

54

, then the number of sides of this polygon is:

HARD

Mathematics>Algebra>Binomial Theorem>Properties of Binomial Coefficients

x = {}^{16}C_{5} + {}^{12}C_{4}, y = \sum_{r = 1}^{3} {}^{(20 - r)}C_{4}, z = \sum_{k = 1}^{4} {}^{(16 - k)}C_{3},

then

x + y + z =

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Mathematics>Algebra>Binomial Theorem>Properties of Binomial Coefficients

Let

(1 + x)^{n} = C_{0} + C_{1} x + C_{2} x^{2} + \dots + C_{n} x^{n},

where

C_{r} = {}^{n}C_{r}

and

(C_{0} + C_{1}) (C_{1} + C_{2}) \dots (C_{n - 1} + C_{n}) = A C_{1} C_{2} \dots C_{n}

, then for

n = 5, A

is equal to

EASY

Mathematics>Algebra>Binomial Theorem>Properties of Binomial Coefficients

In how many ways a team of $5$ members can be chosen from $8$ members?

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Mathematics>Algebra>Binomial Theorem>Properties of Binomial Coefficients

From 6 different novels and 3 different dictionaries, 4 novels and 1 dictionary are to be selected and arranged in a row on the shelf so that the dictionary is always in the middle. Then the number of such arrangement is

If ∑i=1nCi-1nCin+Ci-1n3=3613 then n is equal to

Important Questions on Binomial Theorem

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If $\sum_{i = 1}^{n} {(\frac{{}^{n}C_{i - 1}}{{}^{n}C_{i} + {}^{n}C_{i - 1}})}^{3} = \frac{36}{13}$ then $n$ is equal to