MEDIUM

Earn 100

If $a_{r}$ is the coefficient of $x^{r - 1}$ in ${(1 + x)}^{n} + {(1 + x)}^{n + 1} + . . . + {(1 + x)}^{n + k} (n < r - 1 \leq n + k)$ , then $\sum_{r = 0}^{n + k + 1} {(- 1)}^{r} a_{r}$ is equal to

19.55% studentsanswered this correctly

Important Questions on Binomial Theorem

HARD

Mathematics>Algebra>Binomial Theorem>Series involving Binomial Coefficients

The value of

- {}^{15}C_{1} + 2 \cdot {}^{15}C_{2} - 3 \cdot^{15} C_{3} + . . . . . - 15 \cdot {}^{15}C_{15} + {}^{14}C_{1} + {}^{14}C_{3} + {}^{14}C_{5} + . . . . + {}^{14}C_{11}

is equal to

MEDIUM

Mathematics>Algebra>Binomial Theorem>Series involving Binomial Coefficients

\sum_{r = 1}^{9} {(\frac{r + 3}{2^{r}})}^{9} C_{r} = α {(\frac{3}{2})}^{9} + β,

then

α + β

is equal to

HARD

Mathematics>Algebra>Binomial Theorem>Series involving Binomial Coefficients

n ⩾ 2

is a positive integer, then the sum of the series

{}^{n + 1}C_{2} + 2 ({}^{2}C_{2} + {}^{3}C_{2} + {}^{4}C_{2} + \dots + {}^{n}C_{2})

HARD

Mathematics>Algebra>Binomial Theorem>Series involving Binomial Coefficients

For a complex number $z$ , let $(1 + z)^{15} = a_{0} + a_{1} z + \dots + a_{15} z^{15}$ . The value of

${(a_{0} - 4 a_{2} + 16 a_{4} + \dots - 2^{14} a_{14})}^{2} + {(2 a_{1} - 8 a_{3} + 32 x_{5} \dots - 2^{15} a_{15})}^{2}$ is

HARD

Mathematics>Algebra>Binomial Theorem>Series involving Binomial Coefficients

n

is positive integer and

ω \neq 1

is a cube root of unity, the number of possible values of

|e^{\sum_{k = 0}^{n} (\begin{matrix} n \\ k \end{matrix}) ω^{k}}|

EASY

Mathematics>Algebra>Binomial Theorem>Series involving Binomial Coefficients

(\begin{matrix} 20 \\ 1 \end{matrix}), (\begin{matrix} 20 \\ 2 \end{matrix}), \dots, (\begin{matrix} 20 \\ 20 \end{matrix})

denote binomial coefficients, then

(\begin{matrix} 20 \\ 1 \end{matrix}) - 2 (\begin{matrix} 20 \\ 2 \end{matrix}) + 3 (\begin{matrix} 20 \\ 3 \end{matrix}) - 4 (\begin{matrix} 20 \\ 4 \end{matrix}) + \dots - n (\begin{matrix} 20 \\ 20 \end{matrix})

, is equal to

HARD

Mathematics>Algebra>Binomial Theorem>Series involving Binomial Coefficients

\frac{1}{4} - \frac{5}{4 \cdot 8} + \frac{5 \cdot 9}{4 \cdot 8 \cdot 12} - \dots =

MEDIUM

Mathematics>Algebra>Binomial Theorem>Series involving Binomial Coefficients

{(1 + x)}^{n} = C_{0} + C_{1} x + C_{2} x^{2} + .... + C_{n} x^{n}

, then the value of

C_{0} + 2 C_{1} + 3 C_{2} + ... + (n + 1) C_{n}

MEDIUM

Mathematics>Algebra>Binomial Theorem>Series involving Binomial Coefficients

The sum of the series

{2 .}^{20} C_{0} + {5 .}^{20} C_{1} + {8 .}^{20} C_{2} + {11 .}^{20} C_{3} + . . . . . . . + {62 .}^{20} C_{20}

is equal to

HARD

Mathematics>Algebra>Binomial Theorem>Series involving Binomial Coefficients

^{20} C_{1} + (2^{2})^{20} C_{2} + (3^{2})^{20} C_{3} + . . . . . + (20^{2})^{20} C_{20} = A (2^{β}),

then the ordered pair

(A, β)

is equal to

MEDIUM

Mathematics>Algebra>Binomial Theorem>Series involving Binomial Coefficients

{}^{20}C_{r}

is the co-efficient of

x^{r}

in the expansion of

(1 + x)^{20}

, then the value of

\sum_{r = 0}^{20} r^{2} ({}^{20}C_{r})

is equal to:

MEDIUM

Mathematics>Algebra>Binomial Theorem>Series involving Binomial Coefficients

\sum_{r = 0}^{25} \{({}^{50}C_{r}) ({}^{50 - r}C_{25 - r})\} = K ({}^{50}C_{25})

, then

K

is equal to

MEDIUM

Mathematics>Algebra>Binomial Theorem>Series involving Binomial Coefficients

\sum_{k = 0}^{20} {({}^{20}C_{k})}^{2}

is equal to

MEDIUM

Mathematics>Algebra>Binomial Theorem>Series involving Binomial Coefficients

Let

[x]

denote greatest integer less than or equal to

x

. If for

n \in N, {(1 - x + x^{3})}^{n} = \sum_{j = 0}^{3 n} a_{j} x^{j}

, then

\sum_{j = 0}^{[\frac{3 n}{2}]} a_{2 j} + 4 \sum_{j = 0}^{[\frac{3 n - 1}{2}]} a_{2 j + 1}

is equal to :

HARD

Mathematics>Algebra>Binomial Theorem>Series involving Binomial Coefficients

Let

n > 2

be an integer and define a polynomial

p (x) = x^{n} + a_{n - 1} x^{n - 1} + \dots + a_{1} x + a_{0}

, where

a_{0}, a_{1}, \dots, a_{n - 1}

are integers. Suppose we know that

n p (x) = (1 + x) p^{'} (x)

. If

b = p (1)

, then

MEDIUM

Mathematics>Algebra>Binomial Theorem>Series involving Binomial Coefficients

S = C_{1} + 2 \cdot C_{2} + 3 \cdot C_{3} + \dots + n \cdot C_{n}

, then

S

is equal to

MEDIUM

Mathematics>Algebra>Binomial Theorem>Series involving Binomial Coefficients

C_{0} C_{r} + C_{1} C_{r + 1} + C_{2} C_{r + 2} + \dots + C_{n - r} C_{n}

is equal to

MEDIUM

Mathematics>Algebra>Binomial Theorem>Series involving Binomial Coefficients

The coefficient of

x^{7}

in the expression

{(1 + x)}^{10} + x {(1 + x)}^{9} + x^{2} {(1 + x)}^{8} + . . . . + x^{10}

, is

HARD

Mathematics>Algebra>Binomial Theorem>Series involving Binomial Coefficients

The sum of coefficients of integral powers of

x

in the binomial expansion of

{(1 - 2 \sqrt{x})}^{50}

MEDIUM

Mathematics>Algebra>Binomial Theorem>Series involving Binomial Coefficients

The value of

[C (7, 0) + C (7, 1)] + [C (7, 1) + C (7, 2)] + \dots + [C (7, 6) + C (7, 7)]

If ar is the coefficient of xr-1 in (1+x)n+(1+x)n+1+...+(1+x)n+k n<r-1≤n+k, then ∑r=0n+k+1(-1)rar is equal to

Important Questions on Binomial Theorem

Learn and Prepare for any exam you want !

If $a_{r}$ is the coefficient of $x^{r - 1}$ in ${(1 + x)}^{n} + {(1 + x)}^{n + 1} + . . . + {(1 + x)}^{n + k} (n < r - 1 \leq n + k)$ , then $\sum_{r = 0}^{n + k + 1} {(- 1)}^{r} a_{r}$ is equal to