If N and n be nearly equal, show that N n = N N + n + N + n 4 n very nearly.

Let N = n + h , h is very small that square and above terms can be neglected. Now, LHS = N n 1 2 = n + h n 1 2 = 1 + h n 1 2 1 + 1 2 · h n + … = 1 + h 2 n RHS = N N + n + n + N 4 n ⇒ n + h 2 n + h + 2 n + h 4 n ⇒ n 1 + h n 2 n 1 + h 2 n + 2 n 1 + h 2 n 4 n ⇒ 1 2 1 + h n 1 + h 2 n - 1 + 1 2 1 + h 2 n ⇒ 1 2 1 + h n 1 - h 2 n + 1 2 1 + h 2 n ⇒ 1 2 1 + h n - h 2 n + … + 1 2 1 + h 2 n = 1 2 2 + h n = 1 + h 2 n ∴ Hence, LHS = RHS.

If x is very small in magnitude when compared with a , prove that a a + x 1 2 + a a - x 1 2 = 2 + 3 x 2 4 a 2 nearly.

To prove, a a + x 1 2 + a a - x 1 2 = 2 + 3 x 2 4 a 2 , x < < < 1 1 + x n = 1 + n x + n ( n - 1 ) 2 ! x 2 + n n - 1 n - 2 3 ! x 3 + ⋯ ∞ , n ∈ Q LHS = a a + x 1 2 + a a - x 1 2 = a 1 2 a + x - 1 2 + a - x 1 2 = a 1 2 a - 1 2 1 + x a - 1 2 + 1 - x a - 1 2 = 1 + x a - 1 2 + 1 - x a - 1 2 = 1 - 1 2 x a + 3 8 x 2 a 2 + 1 + 1 2 x a + 3 8 x 2 a 2 = 2 + 3 x 2 4 a 2 = RHS Hence, proved.

If x be nearly equal to 1 , prove that m x m - n x n m - n = x m + n nearly.

To prove, m x m - n x n m - n = x m + n LHS = m x m - n x n m - n ...(i) Since x is nearly equal to 1 Let x = 1 + h h < < < 1 ...(ii) Substituting (i) in (ii) LHS = m ( 1 + h ) m - n ( 1 + h ) n m - n We know that, 1 + x n = 1 + n x + n ( n - 1 ) 2 ! x 2 + n ( n - 1 ) ( n - 2 ) 3 ! x 3 + ⋯ ∞ ; | x | < 1 and n is any real number. ⇒ LHS = m ( 1 + m h ) - n ( 1 + n h ) m - n ...since h < < < 1 = m - n + m 2 - n 2 h m - n = 1 + m 2 - n 2 h m - n = 1 + ( m + n ) h = ( 1 + h ) m + n = x m + n = RHS Hence, proved.

Prove that 1 + x + x 2 + … ∞ 2 = 1 + 2 x + 3 x 2 + … ∞ .

To prove, 1 + x + x 2 + ⋯ ∞ 2 = 1 + 2 x + 3 x 2 + ⋯ ∞ LHS = 1 + x + x 2 + ⋯ ∞ 2 From standard binomial expansions of negative indices, we know that ( 1 - x ) - 1 = 1 + x + x 2 + ⋯ · ∞ ⇒ LHS = ( 1 - x ) - 1 2 = ( 1 - x ) - 2 1 + x n = 1 + n x + n ( n - 1 ) 2 ! x 2 + n n - 1 n - 2 3 ! x 3 + ⋯ ∞ , n ∈ Q and x < 1 ⇒ LHS = 1 + 2 x + ( - 2 ) ( - 3 ) 2 ! x 2 + ⋯ ∞ = 1 + 2 x + 3 x 2 + ⋯ ∞ = RHS Hence, proved.

Prove that 1 + x + x 2 + … ∞ 3 = 1 + 3 x + 6 x 2 + … ∞ .

To prove, 1 + x + x 2 + ⋯ ∞ 3 = 1 + 3 x + 6 x 2 + ⋯ ∞ LHS = 1 + x + x 2 + ⋯ ∞ 3 From standard binomial expansions of negative indices, we know that ( 1 - x ) - 1 = 1 + x + x 2 + ⋯ · ∞ ⇒ LHS = ( 1 - x ) - 1 3 = ( 1 - x ) - 3 1 + x n = 1 + n x + n ( n - 1 ) 2 ! x 2 + n n - 1 n - 2 3 ! x 3 + ⋯ ∞ , n ∈ Q and x < 1 ⇒ LHS = 1 + 3 x + ( - 3 ) ( - 4 ) 2 x 2 + ⋯ ∞ = 1 + 3 x + 6 x 2 + ⋯ ∞ = RHS Hence, proved.

Prove that, 1 + x + x 2 + … ∞ 1 + 2 x + 3 x 2 + … ∞ = 1 + 3 x + 6 x 2 + … ∞ .

To prove, 1 + x + x 2 + … ∞ 1 + 2 x + 3 x 2 + … ∞ = 1 + 3 x + 6 x 2 + … ∞ LHS = 1 + x + x 2 + … ∞ 1 + 2 x + 3 x 2 + … ∞ From standard binomial expansions of negative indices, we know that ( 1 - x ) - 1 = 1 + x + x 2 + ⋯ · ∞ and ( 1 - x ) - 2 = 1 + 2 x + 3 x 2 + ⋯ ⋯ ∞ ⇒ LHS = 1 + x + x 2 + … ∞ 1 + 2 x + 3 x 2 + … ∞ = ( 1 - x ) - 1 ( 1 - x ) - 2 = ( 1 - x ) - 3 1 + x n = 1 + n x + n ( n - 1 ) 2 ! x 2 + n n - 1 n - 2 3 ! x 3 + ⋯ ∞ , n ∈ Q and x < 1 ⇒ LHS = 1 + 3 x + ( - 3 ) ( - 4 ) 2 ! x 2 + ⋯ ∞ = 1 + 3 x + 6 x 2 + ⋯ ∞ = RHS Hence, proved.

If the binomial expansion of ( a + b x ) - 2 is 1 4 - 3 x + … , find the values of a and b .

Given, ( a + b x ) - 2 = 1 4 - 3 x + … ⇒ 1 a 2 1 + b x a - 2 1 + x n = 1 + n x + n ( n - 1 ) 2 ! x 2 + n n - 1 n - 2 3 ! x 3 + ⋯ ∞ , n ∈ Q and x < 1 Now we expand the series 1 + b x a - 2 = 1 + - 2 b x a + - 2 ( - 2 - 1 ) 2 ! b x a 2 + ⋯ ∞ , n ∈ Q ∴ 1 a 2 1 + b x a - 2 = 1 a 2 ( 1 + - 2 b x a + - 2 ( - 2 - 1 ) 2 ! b x a 2 + ⋯ ∞ ) = 1 4 - 3 x + . . . Equating corresponding coefficients. ⇒ 1 a 2 = 1 4 ⇒ a = 2 ⇒ - 2 b a 3 = - 3 ⇒ b = 12 Hence, a = 2 , b = 12 .

E M B I B E

WBCHSE Mathematics for Class 11>Algebra>Chapter 8 - Binomial Theorem>EXERCISE 6(B)

R S Aggarwal Solutions for Exercise 2: EXERCISE 6(B)

Author:R S Aggarwal

R S Aggarwal Mathematics Solutions for Exercise - R S Aggarwal Solutions for Exercise 2: EXERCISE 6(B)

Attempt the practice questions from Exercise 2: EXERCISE 6(B) with hints and solutions to strengthen your understanding. WBCHSE Mathematics for Class 11 solutions are prepared by Experienced Embibe Experts.

Questions from R S Aggarwal Solutions for Exercise 2: EXERCISE 6(B) with Hints & Solutions

EASY

11th West Bengal Board

IMPORTANT

WBCHSE Mathematics for Class 11>Chapter 8 - Binomial Theorem>EXERCISE 6(B)>Q 30

If $N$ and $n$ be nearly equal, show that $\sqrt{\frac{N}{n}} = \frac{N}{N + n} + \frac{N + n}{4 n}$ very nearly.

MEDIUM

11th West Bengal Board

IMPORTANT

WBCHSE Mathematics for Class 11>Chapter 8 - Binomial Theorem>EXERCISE 6(B)>Q 31

If $x$ is very small in magnitude when compared with $a$ , prove that ${(\frac{a}{a + x})}^{\frac{1}{2}} + {(\frac{a}{a - x})}^{\frac{1}{2}} = (2 + \frac{3 x^{2}}{4 a^{2}})$ nearly.