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Explore popular questions from Binomial Theorem for JEE Advanced. This collection covers Binomial Theorem previous year JEE Advanced questions hand picked by experienced teachers.

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Binomial Theorem

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Q 1. Let {tex} T _ { 1 } = 7 , T _ { 2 } = 7 ^ { 7 } , T _ { 3 } = 7 ^ { 7 ^ { 7 } } {/tex} and so on. The digit at the tens places of number {tex} T _ { 1000 } {/tex} is

A

8

B

0

C

6

4

Explanation


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Q 2. If the polynomial {tex} f ( x ) = 1 - x + x ^ { 2 } - x ^ { 3 } + \cdots + x ^ { 19 } + x ^ { 20 } {/tex} is expressed as {tex} g ( y ) = a _ { 0 } + a _ { 1 } y + a _ { 2 } y ^ { 2 } + \cdots + a _ { 20 } y ^ { 20 } , {/tex} where {tex} y = x - 4 {/tex}, then the value of {tex} a _ { 0 } + a _ { 1 } + a _ { 2 } + \cdots + a _ { 20 } {/tex} is

A

{tex} \frac { 5 ^ { 21 } - 1 } { 6 } {/tex}

B

{tex} \frac { 5 ^ { 20 } } { 6 } {/tex}

C

{tex} \frac { 1 + 5 ^ { 20 } } { 6 } {/tex}

{tex} \frac { 1 + 5 ^ { 21 } } { 6 } {/tex}

Explanation

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Q 3. {tex} ^{2 n + 3} C _ { 1 } + ^ { 2 n + 3 } C _ { 2 } + \cdots + ^ { 2 n + 3 } C _ { n } - ^{2 n + 3} C_{2n+3}- ^{2 n + 3} C_{2n+2}-\cdots- ^{2 n + 3} C_{n+3} {/tex} is equal to

A

{tex} a {/tex}

B

{tex} ^{2 n + 3} C _ { n + 1 } {/tex}

{tex} - 1 {/tex}

D

0

Explanation

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Q 4. If {tex} ( 1 + x ) ^ { n } = ^ { n } C _ { 0 } + ^ { n } C _ { 1 } x + ^ { n } C _ { 2 } x ^ { 2 } + \cdots + ^ { n } C _ { n } x ^ { n } {/tex} where {tex} ^ { n } C _ { 0 }, ^ { n } C _ { 1 }, ^{ n } C _ { 2 } , \cdots {/tex} are binomial coefficients. Then {tex} 2 \left( C _ { 0 } + C _ { 3 } + C _ { 6 } + \cdots \right) + \left( C _ { 1 } + C _ { 4 } \right. {/tex} {tex} \left. + C _ { 7 } + \cdots \right) ( 1 + \omega ) + \left( C _ { 2 } + C _ { 5 } + C _ { 8 } + \cdots \right) \left( 1 + \omega ^ { 2 } \right) , {/tex} where {tex} \omega {/tex} is the cube root of unity and {tex} n {/tex} is a multiple of {tex} 3 , {/tex} is equal to

A

{tex} 2 ^ { n } + 1 {/tex}

B

{tex} 2 ^ { n - 1 } + 1 {/tex}

C

{tex} 2 ^ { n + 1 } - 1 {/tex}

{tex} 2 ^ { n } - 1 {/tex}

Explanation


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Q 5. If {tex} b _ { 1 } = 2 {/tex} and {tex} b _ { n } = n \left( 1 + b _ { n - 1 } \right) \forall n \geq 2 , {/tex} then {tex} \displaystyle \lim _ { n \rightarrow \infty } \frac { b _ { n + 2 } } { ( n + 2 ) ! } {/tex} is equal to

A

{tex} e {/tex}

B

{tex} 2 e {/tex}

C

{tex} e - 1 {/tex}

{tex} e + 1 {/tex}

Explanation

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Q 6. If {tex} ( x + 1 ) ( x + 2 ) ( x + 3 ) \cdots ( x + n ) = A _ { 0 } + A _ { 1 } x + A _ { 2 } x ^ { 2 } + \cdots + A _ { n } x ^ { n } {/tex}, then {tex} A _ { 1 } + 2 A _ { 2 } + \cdots + n A _ { n } {/tex} is equal to

A

{tex} ( n - 1 ) ! \left( 1 + \frac { 1 } { 2 } + \frac { 1 } { 3 } + \cdots + \frac { 1 } { n + 1 } \right) {/tex}

B

{tex} \left( 1 + \frac { 1 } { 2 } + \frac { 1 } { 3 } + \cdots + \frac { 1 } { n + 1 } \right) {/tex}

{tex} ( n + 1 ) ! \left( \frac { 1 } { 2 } + \frac { 1 } { 3 } + \cdots + \frac { 1 } { n + 1 } \right) {/tex}

D

None of these

Explanation



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Q 7. {tex}\displaystyle \sum _ { r = 0 } ^ { 10 } 2 ^ { 10 } \ {^{20} C _ { r }} {^ { 20 - r } C _ { 10 - r }} {/tex} is equal to

A

{tex} ^ { 20 } C _ { 10 } {/tex}

B

{tex} ^ { 20 } C _ { 10 } \left( \frac { 3 } { 2 } \right) ^ { 10 } {/tex}

{tex} ^ { 20 } C _ { 10 } \ 3 ^ { 10 } {/tex}

D

{tex} ^ { 20 } C _ { 10 }\ 2 ^ { 10 } {/tex}

Explanation