Show that C 0 + 3 C 1 + 5 C 2 + … . + ( 2 n + 1 ) C n = ( n + 1 ) 2 n .

Consider the following series 1 + x 2 n = 1 + C 1 n x 2 + C 2 n x 4 + C 3 n x 6 + … + C n n x 2 n Multiplying with x on both sides ⇒ x 1 + x 2 n = x + C 1 n x 3 + C 2 n x 5 + C 3 n x 7 + … + C n n x 2 n + 1 Now differentiating both sides with respect to x and plug in x = 1 . 1 + x 2 n + n x ( 2 x ) 1 + x 2 n - 1 x = 1 = 1 + 3 n C 1 + 5 n C 2 + … + 2 n + 1 n C n ⇒ 1 + 3 n C 1 + 5 n C 2 + … + ( 2 n + 1 ) C n n = 1 + x 2 n + n x ( 2 x ) 1 + x 2 n - 1 x = 1 = 2 n + 2 n 2 n - 1 = 2 n + n 2 n = 2 n ( n + 1 ) Hence, 1 + 3 n C 1 + 5 n C 2 + … ( 2 n + 1 ) C n n = 2 n ( n + 1 ) .

Show that C 0 + 3 C 1 + 5 C 2 + … . + ( 2 n + 1 ) C n = ( n + 1 ) 2 n .

Consider the following series 1 + x 2 n = 1 + C 1 n x 2 + C 2 n x 4 + C 3 n x 6 + … + C n n x 2 n Multiplying with x on both sides ⇒ x 1 + x 2 n = x + C 1 n x 3 + C 2 n x 5 + C 3 n x 7 + … + C n n x 2 n + 1 Now differentiating both sides with respect to x and plug in x = 1 . 1 + x 2 n + n x ( 2 x ) 1 + x 2 n - 1 x = 1 = 1 + 3 n C 1 + 5 n C 2 + … + 2 n + 1 n C n ⇒ 1 + 3 n C 1 + 5 n C 2 + … + ( 2 n + 1 ) C n n = 1 + x 2 n + n x ( 2 x ) 1 + x 2 n - 1 x = 1 = 2 n + 2 n 2 n - 1 = 2 n + n 2 n = 2 n ( n + 1 ) Hence, 1 + 3 n C 1 + 5 n C 2 + … ( 2 n + 1 ) C n n = 2 n ( n + 1 ) .

Find the sum of 1 · 2 C 2 + 2 · 3 C 3 + … . + ( n - 1 ) n C n .

We know the series 1 + x n = C 0 + C 1 x + C 2 x 2 + C 3 x 3 + · · · Differentiate both sides with respect to x . n ( 1 + x ) n - 1 = C 1 + 2 C 2 x + 3 C 3 x 2 + . . . + n C n x n - 1 Differentiate again with respect to x . n n - 1 ( 1 + x ) n - 2 = 1 . 2 C 2 x + 2 . 3 C 3 x 2 + . . . + n - 1 n C n x n - 2 Plug in x = 1 . n n - 1 2 n - 1 = 1 . 2 C 2 + 2 . 3 C 3 + . . . + n - 1 n C n Hence, 1 . 2 C 2 + 2 . 3 C 3 + . . . + n - 1 n C n = n n - 1 2 n - 1

Find the sum of C 1 + 2 C 2 2 + 3 2 C 3 + … . + n 2 C n .

We know the standard series C 0 + C 1 x + C 2 x 2 + … + C n x n = ( 1 + x ) n Differentiating with respect to x . 0 + C 1 + 2 C 2 x + … + n C n x n - 1 = n ( 1 + x ) n - 1 Multiply with x on both sides. C 1 x + 2 C 2 x 2 + … + n C n x n = n x ( 1 + x ) n - 1 Again differentiate with respect to x . C 1 + 2 2 C 2 x + 3 2 C 3 x 2 + . … + n 2 C n x n = n ( 1 + x ) n - 1 + n x n - 1 ( 1 + x ) n - 2 Plug in x = 1 . C 1 + 2 2 C 2 + 3 2 C 3 + . … + n 2 C n = n ( 1 + 1 ) n - 1 + n . 1 . n - 1 ( 1 + 1 ) n - 2 C 1 + 2 2 C 2 x + 3 2 C 3 x 2 + . … + n 2 C n x n = n ( 2 ) n - 1 + n n - 1 ( 2 ) n - 2 Hence, C 1 + 2 2 C 2 x + 3 2 C 3 x 2 + . … + n 2 C n x n = n n + 1 2 n - 2

Find the sum of C 0 - 1 2 C 1 + 1 3 C 2 - … + - 1 n 1 n + 1 C n .

Given, C 0 - 1 2 C 1 + 1 3 C 2 + ⋯ + ( - 1 ) n 1 n + 1 C n = 1 - 1 2 n + 1 3 n ( n - 1 ) 2 ! + ⋯ + - 1 n 1 n + 1 C n Taking 1 n + 1 term as common and write it outside. = 1 n + 1 ( n + 1 ) - n ( n + 1 ) 2 ! + ( n + 1 ) n ( n - 1 ) 3 ! + … + ( - 1 ) n Adding and subtracting with 1 . = 1 n + 1 - 1 + ( n + 1 ) - n ( n + 1 ) 2 ! + ⋯ + ( - 1 ) n + 1 = - 1 n + 1 1 - ( n + 1 ) + n ( n + 1 ) 2 ! - . . . + ( - 1 ) n - 1 = - 1 n + 1 0 - 1 = 1 n + 1 Hence, C 0 - 1 2 C 1 + 1 3 C 2 - … + ( - 1 ) n 1 n + 1 C n = 1 n + 1 .

Show that C 1 2 + 2 C 2 2 + 3 C 3 2 + … . + n C n 2 = ( 2 n - 1 ) ! { ( n - 1 ) ! } 2 .

We know standard series 1 + x n = C 0 + C 1 x + C 2 x 2 + C 3 x 3 + ⋯ + C n x n · · · 1 x + 1 n = C n + C n - 1 x + C n - 2 x 2 + ⋯ + C 1 x n - 1 + C 0 x n · · · 2 Differentiate with respect to x on both sides n ( 1 + x ) n - 1 = C 1 + 2 C 2 x + 3 C 3 x 2 + ⋯ + n C n x n - 1 Multiplying with x on both sides n x ( 1 + x ) n - 1 = C 1 x + 2 C 2 x 2 + 3 C 3 x 3 + ⋯ + n C n x n Multiplying both equations 1 and 2 n x 1 + x n - 1 × 1 + x n = 0 + C 1 x + 2 C 2 x 2 + 3 C 3 x 3 + ⋯ + n C n x n C n + C n - 1 x + C n - 2 x 2 + ⋯ + C 1 x n - 1 + C 0 x n Comparing coefficients of x n on both sides n × C n - 1 2 n - 1 x n = x n C 1 2 + 2 C 2 2 + 3 C 3 2 + … + n C n 2 C 1 2 + 2 C 2 2 + 3 C 3 2 + … + n C n 2 = n × C n - 1 2 n - 1 C 1 2 + 2 C 2 2 + 3 C 3 2 + … + n C n 2 = n × 2 n - 1 ! n - 1 ! 2 n - 1 - n - 1 ! C 1 2 + 2 C 2 2 + 3 C 3 2 + … + n C n 2 = n 2 n - 1 ! n - 1 ! n n - 1 ! Hence, C 1 2 + 2 C 2 2 + 3 C 3 2 + · · · + n C n 2 = ( 2 n - 1 ) ! { ( n - 1 ) ! } 2

Show that C 2 + 2 C 3 + 3 C 4 + … . . + ( n - 1 ) C n = 1 + ( n - 2 ) 2 n - 1 .

( 1 + x ) n = C 0 + C 1 x + C 2 x 2 + ⋯ + C n - 1 x n - 1 + c n x n Dividing with x on both sides. ( 1 + x ) n x = C 0 x + C 1 + C 2 x + C 3 x 2 + ⋯ + C n - 1 x n - 2 + C n x n - 1 Differentiate with respect to x on both sides. x n ( 1 + x ) n - 1 - 1 + x n x n = - C 0 x 2 + C 2 + 2 C 3 x + . . . + n - 2 C n - 2 x n - 3 + n - 1 C n x n - 2 x n ( 1 + x ) n - 1 - ( 1 + x ) n x n + C 0 x 2 = C 2 + 2 C 3 x + 3 C 4 x 2 + ⋯ + n - 1 C n x n - 2 Plug in x = 1 . n 2 n - 1 - 2 n + 1 = C 2 + 2 C 3 + 3 C 4 + ⋯ + ( n - 1 ) C n C 2 + 2 C 3 + 3 C 4 + ⋯ + ( n - 1 ) C n = 1 + ( n - 2 ) 2 n - 1 LHS = RHS

Prove that C 1 - 1 2 C 2 + 1 3 C 3 - … … … + - 1 n + 1 1 n C n = 1 + 1 2 + … + 1 n .

We have the standard series, ( 1 + x ) n = C 0 + C 1 x + C 2 x 2 + ⋯ + C n x n Dividing with x on both sides. ( 1 + x ) n x = C 0 x + C 1 1 + C 2 x + ⋯ + C n x n - 1 ( 1 + x ) n x - 1 x = C 1 + C 2 x + … + C n x n - 1 1 x ( 1 + x ) n - 1 = C 1 + C 2 x + … + C n x n - 1 Expanding Left hand side of the equation in G.P Series where a = 1 , r = 1 + x . 1 + ( 1 + x ) + ( 1 + x ) 2 + ⋯ + ( 1 + x ) n - 1 = C 1 + C 2 x + ⋯ + C n x n - 1 Now, integrating with respect to x from - 1 to 0 . ∫ - 1 0 ( 1 + ( 1 + x ) + ( 1 + x ) 2 + ⋯ + ( 1 + x ) n - 1 ) d x = ∫ - 1 0 ( C 1 + C 2 x + ⋯ + C n x n - 1 ) d x x + ( 1 + x ) 2 2 + ( 1 + x ) 3 3 + ⋯ + ( 1 + x ) n n - 1 0 = C 1 x + C 2 x 2 2 + ⋯ + C n x n n - 1 0 0 + 1 2 + 1 3 + ⋯ + 1 n - [ - 1 ] = 0 + C 1 - 1 2 C 2 + 1 3 C 3 - … … … + ( - 1 ) n + 1 1 n C n Hence proved that, C 1 - 1 2 C 2 + 1 3 C 3 - … … … + ( - 1 ) n + 1 1 n C n = 1 + 1 2 + … + 1 n .

Prove that C 0 C 1 + C 1 C 2 + … . + C n - 1 C n = 2 n · n · 1 · 3 · 5 … ( 2 n - 1 ) ( n + 1 ) ! .

We know that ( 1 + a ) n = C 0 + C 1 a + C 2 a 2 + … . + C n a n We can also write above equation as follows ( a + 1 ) n = C 0 a n + C 1 a n - 1 + C 2 a n - 2 + … . + C n Multiplying above two equations, we get ( 1 + a ) 2 n = C 0 + C 1 a + C 2 a 2 + … … + C n a n C 0 a n + C 1 a n - 1 + C 2 a n - 2 + … . + C n But ( 1 + a ) 2 n can be written as ( 1 + a ) 2 n = C 0 2 n + C 1 2 n a + C 2 2 n a 2 + … . + C n - 1 2 n a n - 1 + C n 2 n a n + C n + 1 2 n a n + 1 + … . + a 2 n Equating both ( 1 + a ) 2 n terms and equating a n - 1 terms, we get C 0 C 1 + C 1 C 2 + C 2 C 3 + … … + C n - 1 C n = C n - 1 2 n = 2 n ! n - 1 ! n + 1 ! = 2 n ( 2 n - 1 ) ( 2 n - 2 ) . . . 3 , 2 , 1 ( n - 1 ) ! ( n + 1 ) ! = 2 n n ( n - 1 ) ( n - 2 ) … 2 , 1 ( 2 n - 1 ) ( 2 n - 3 ) . . . 5 , 3 , 1 ( n - 1 ) ! ( n + 1 ) ! = 2 n · n · ( n - 1 ) ! 1 · 3 · 5 · ( 2 n - 1 ) ( n - 1 ) ! ( n + 1 ) ! = 2 n · n · 1 · 3 · 5 … ( 2 n - 1 ) ( n + 1 ) ! Hence, C 0 C 1 + C 1 C 2 + … + C n - 1 C n = 2 n · n · 1 · 3 · 5 … ( 2 n - 1 ) ( n + 1 ) ! .

E M B I B E

Bureau's Higher Secondary Elements of Mathematics Vol.1>Chapter 9 - Binomial Theorem>EXERCISES - 9 (b)>Q 4

EASY

11th Odisha Board

IMPORTANT

Earn 100

Show that $C_{0} + 3 C_{1} + 5 C_{2} + \dots . + (2 n + 1) C_{n} = (n + 1) 2^{n}$ .

Important Questions on Binomial Theorem

EASY

11th Odisha Board

IMPORTANT

Bureau's Higher Secondary Elements of Mathematics Vol.1>Chapter 9 - Binomial Theorem>EXERCISES - 9 (b)>Q 5

Find the sum of $C_{1} - 2 C_{2} + 3 C_{3} - \dots . + n (- 1)^{n - 1} C_{n}$ .

EASY

11th Odisha Board

IMPORTANT

Bureau's Higher Secondary Elements of Mathematics Vol.1>Chapter 9 - Binomial Theorem>EXERCISES - 9 (b)>Q 5

Find the sum of $1 \cdot 2 C_{2} + 2 \cdot 3 C_{3} + \dots . + (n - 1) n C_{n}$ .

EASY

11th Odisha Board

IMPORTANT

Bureau's Higher Secondary Elements of Mathematics Vol.1>Chapter 9 - Binomial Theorem>EXERCISES - 9 (b)>Q 5

Find the sum of $C_{1} + 2 {}^{2}C_{2} + 3^{2} C_{3} + \dots . + n^{2} C_{n}$ .

MEDIUM

11th Odisha Board

IMPORTANT

Bureau's Higher Secondary Elements of Mathematics Vol.1>Chapter 9 - Binomial Theorem>EXERCISES - 9 (b)>Q 5

Find the sum of $C_{0} - \frac{1}{2} C_{1} + \frac{1}{3} C_{2} - \dots + {(- 1)}^{n} \frac{1}{n + 1} C_{n}$ .

MEDIUM

11th Odisha Board

IMPORTANT

Bureau's Higher Secondary Elements of Mathematics Vol.1>Chapter 9 - Binomial Theorem>EXERCISES - 9 (b)>Q 6

Show that $C_{1}^{2} + 2 C_{2}^{2} + 3 C_{3}^{2} + \dots . + n C_{n}^{2} = \frac{(2 n - 1)!}{{(n - 1)!}^{2}}$ .

MEDIUM

11th Odisha Board

IMPORTANT

Bureau's Higher Secondary Elements of Mathematics Vol.1>Chapter 9 - Binomial Theorem>EXERCISES - 9 (b)>Q 6

Show that $C_{2} + 2 C_{3} + 3 C_{4} + \dots . . + (n - 1) C_{n} = 1 + (n - 2) 2^{n - 1}$ .

HARD

11th Odisha Board

IMPORTANT

Bureau's Higher Secondary Elements of Mathematics Vol.1>Chapter 9 - Binomial Theorem>EXERCISES - 9 (b)>Q 7

Prove that $C_{1} - \frac{1}{2} C_{2} + \frac{1}{3} C_{3} - \dots \dots \dots + {(- 1)}^{n + 1} \frac{1}{n} C_{n} = 1 + \frac{1}{2} + \dots + \frac{1}{n}$ .