Exponential and logarithm series

Exponential Series

e^x = 1 + x1! + x²2! + x³3! + …… to ∞ This is called exponential series.

Some Particular Cases

1. e = 1 + 11! + 12! + 13! + …… ∞
2. e^-1 = 1 - 11! + 12! - 13! + …… ∞
3. e^-x = 1 - x1! + x²2! - x³3! + …… to ∞
4. If a > 0. a^x = e^{x log a} = 1 + x log a1! + (x log a)²2! + …… to ∞

Illustration 1: ∑_n=1^∞1(2n-1)! equals

(A) e - e^-12

(B) e + e^-12

(C) e^-1 - e²2

(D) none of these

Solution: (B). ∑_n=1^∞1(2n-1)! = 11! + 13! + 15! + …∞ = e - e^-12

Illustration 2: If e^5x + e^xe^3x is expanded in the series of ascending powers of x and n is odd natural number, then the coefficient of xⁿ is

(A) 2ⁿn!

(B) 2ⁿ⁺¹2n!

(C) 2²ⁿ2n!

(D) none of these

Solution: (D).e^5x + e^xe^3x = e^2x + e^-2x = 2(1 + 2²x²2! + 2⁴x⁴4! + ……)

This expansion does not contain any odd power of x ⇒ coefficient of xⁿ when n is odd natural number is zero.
Exercise 1:Ques. ∑_n=1^∞1(n-1)! is equal to

(A) e

(B) e - 1

(C) e - 2

(D) e - 3

Ques. ∑_n=2^∞n²n! is equal to

(A) 2e

(B) 3e

(C) e/2

(D) none of these

Ques. If S = 11.2 + 12.3 + 13.4 + 14.5 + ……, then e^S equals

(A) log_e (4/e)

(B) 4/e

(C) log_e (e/4)

(D) e/4

Logarithmic Series

log_e(1 + x) = x - x²2 + x³3 - x⁴4 + …… to ∞. This is called logarithmic series.

Some Particular Cases

1. Replacing x by -x, |x| < 1: log_e(1 - x) = -x - x²2 - x³3 - x⁴4 - …… to ∞
2. log_e1 + x1 - x = 2(x + x³3 + x⁵5 + …… to ∞)

Illustration 3: The sum of the series log₄ 2 - log₈ 2 + log₁₆ 2 - ... is

(A) e²

(B) log_e 2 + 1

(C) log_e 3 - 2

(D) 1 - log_e 2

Solution: (D). given summation = 1log₂ 4 - 1log₂ 8 + 1log₂ 16 - 1log₂ 32 + ……

= 12 - 13 + 14 - 15 + …… = 1 - (11 + 12 + 13 + 14 + 15 + ……) = 1 - log_e 2

Illustration 4: The sum of the series log_e n + (log_e n)³3 + (log_e n)⁵5 + ..... upto ∞ is

(A) n² - 12n

(B) n² + 12n

(C) n(n + 1)2n

(D) n(n - 1)2n

Solution: (A). Sum of the given series = z + z³3 + z⁵5 + ……, where z = log_e n = 12(e^{log_e n} - e^{-log_e n})

= 12(n - 1n) = n² - 12n

Exercise 2:Ques. 2[m - nm + n + 13(m - nm + n)³ + 15(m - nm + n)⁵ + ……] is equal to

(A) log(mn)

(B) log(nm)

(C) log mn

(D) none of these

Ques. If S = ∑_n=0^∞(log_e x)²ⁿ(2n)!, then S equals to(

A) x + 1x

(B) x - 1x

(C) 12(x + 1x)

(D) 12(x - 1x)

Ques. The sum of the series 12·2² + 13·2³ + 14·2⁴ + …… is

(A) log_e(2/3)

(B) log₁₀(3/2)

(C) log_e(3/2)

(D) none of these

Answer to ExercisesExercise 1

(i) A

(ii) A

(iii) A

Exercise 2

(i) A

(ii) C

(iii) C

Formulae and Concepts at a Glance

1. e^x = 1 + x + x²2! + x³3! + …∞ holds for all real values of x.
2. log(1 + x) = x - x²2 + x³3 - x⁴4 + … is valid for -1 < x ≤ 1.
3. log(1 - x) = -x - x²2 - x³3 - x⁴4 - …
4. log 2 = 1 - 12 + 13 - 14 + …∞
5. a^x = 1 + x log_e a + (x log_e a)²2! + …∞

Solved Examples

Ques 1. The coefficient of xⁿ in the expansion of e^{a + bx} is

(A) e^abⁿn!

(B) e^ab^-nn!

(C) e^baⁿn!

(D) none of these

Sol. (A). e^{a + bx} = e^a·e^bx = e^a ∑_n=0^∞(bx)ⁿn! ∴ coefficient of xⁿ = e^abⁿn!

Ques 2. Sum to n terms of the series 92! + 163! + 274! + 425! + …… is

(A) 11e - 6

(B) 11e + 6

(C) 6e - 11

(D) none of these

Sol. (A). Let T_n denote the nth term of the given series, then T_n = a_nn!, where a_n denotes then the term of the series 9 + 16 + 27 + 42 + ……., in which differences from an A.P.

To find a_n: Let s_n be the sum of first n terms of the series 9 + 16 + 27 + 42 + ….., then

s_n = 9 + 16 + 27 + 42 + ……+ a_n-1 + a_n ….(1)

Again s_n = 9 + 16 + 27 +…..+ a_n-1 + a_n ….(2)

Subtracting (2) from (1), we obtain 0 = 9 + 7 + 11 + 15 +….. upto n terms - a_n

∴ a_n = 9 + {7 + 11 + 15 + …… upto (n - 1) terms}

= 9 + (n-1)2[2×7 + (n-1-1)×4] = 9 + (n-1)2[14 + 4n - 8]

= 9 + (n-1)(6 + 4n)2 = 9 + (n - 1)(3 + 2n) = 9 + 2n² + n - 3 = 2n² + n + 6

Hence, T_n = 2n² + n + 6n! = 2n²n! + nn! + 6n!

= 2n(n-1)! + 1(n-1)! + 6n!

= 2(n-2)! + 2(n-1)! + 1(n-1)! + 6n!

= 2(n-2)! + 3(n-1)! + 6n!

Substituting n = 2, 3, 4,…. and adding, we get T₂ + T₃ + T₄ + ….. to ∞;

= 2(10! + 11! + 12! + ……) + 3(11! + 12! + 13! + ……) + 6(12! + 13! + 14! + ……)

= 2e + 3(e - 1) + 6(e - 2); Adding T₁ to both sides, we get

Sum of the given series = 11e - 15 + T₁; = 11e - 15 + 9 = 11e - 6.

Ques 3. The sum of the series 1²·21! + 2²·32! + 3²·43! + 4²·54! + …… is equal to

(A) 5e

(B) 3e

(C) 7e

(D) 2e

Sol. (C). Given series is ∑_n=1^∞n²(n+1)n!

Let T_n denote the nth term, then T_n = n²(n+1)n! = n²(n+1)n! = n+1(n-1)! = n+1(n-1)!

= n(n-1)! + 1(n-1)! = 1(n-2)! + 1(n-1)!

= 1(n-2)! + 1(n-1)! = 2(n-2)! + 3(n-1)! + 1(n-2)! + 4(n-1)!

Ques 4. 12! + 1+23! + 1+2+34! + …… is equal to

(A) e2

(B) e3

(C) e4

(D) e5

Sol. (A). Here, T_n = 1+2+3+….+n(n+1)! = n(n+1)/2(n+1)! = n(n+1)2(n+1)! = 12(n-1)!

put n = 1, 2, 3,…. and add to obtain the required sum = e/2

Ques 5. If e^x1-x = B₀ + B₁x + B₂x² + …+ B_nxⁿ + …, then B_n - B_n-1 equals

(A) 1n!

(B) 1(n-1)!

(C) 1n! - 1(n-1)!

(D) 1

Sol. (A). e^x = (B₀ + B₁x + B₂x² + …+ B_nxⁿ + …)(1 - x)

= (B₀ + B₁x + B₂x² + …+ B_nxⁿ + …) - (B₀x + B₁x² + B₂x³ + …+ B_n-1xⁿ + …)

∑_n=0^∞xⁿn! = B₀ + ∑_n=1^∞ (B_n - B_n-1)xⁿ

equating coeff. of xⁿ on both sides, B_n - B_n-1 = 1n!

Q6. If a_n = n(n+1)2n!, then the sum of the series ∑a_n is

(A) e

(B) e⁻¹

(C) 3e2

(D) e2

Sol. (C). a_n = n(n+1)2n! = 12(n+1(n-1)!) = 12(1(n-2)! + 1(n-1)!)

∑a_n = 12∑(1(n-2)! + 1(n-1)!) = 12(e + e) = 3e2

Ques 7. x²2 + 23x³ + 34x⁴ + 45x⁵ + … is

(A) x1+x

(B) log(1 + x)

(C) x1-x + log(1 + x)

(D) none of these

Sol. (B). T_n = nn+1xⁿ⁺¹ = xⁿ⁺¹ - xⁿ⁺¹n+1; Putting n = 1, 2, 3, …

S = (x² + x³ + x⁴ + …) - (x²2 + x³3 + x⁴4 + …) = x1-x + log(1-x)

Ques 8. If 0 < y ≤ 2^1/3 and x(y³ - 1) = 1, then 2x + 23x³ + 25x⁵ + …… is equal to

(A) 13log(y³2-y³)

(B) 13log(y³1-y³)

(C) 13log(2y³1-y³)

(D) 13log(y³1-2y³)

Sol. (A).2x + 23x³ + 25x⁵ + ……… = 2t + 2t³ + 2t⁵ + ……

where t = 1x = y³ - 1

= 2 × 12log(1+t1-t) = log(1+y³-11-y³+1) = log(y³2-y³) = 13log(y³2-y³)

Ques 9. The sum of the series log₄2 - log₈2 + log₁₆2 + … is

(A) e²

(B) log_e2 + 1

(C) log_e3 - 2

(D) 1 - log_e2

Sol. (D). Given series = 1log₂4 - 1log₂8 + 1log₂16 - 1log₂32 + ……

= 12 - 13 + 14 - 15 + ……

= 1 - (11 + 12 + 13 + 14 + 15 + ……) = 1 - log_e2

Ques 10. Coefficient of x⁴ in the expansion of 1-3x+x²e^x is

(A) 2524

(B) 2425

(C) 425

(D) 524

Sol. (A). (1 - 3x + x²) e^-x

= (1 - 3x + x²)(1 - x + x²2! - x³3! + x⁴4! - ……)

Coefficient of x⁴ = 14! + 13! + 12! = 2524

Assignment Problems

Ques. If e^x = y + √(1 + y²), then the value of y is

(A) e^x - e^-x

(B) 12(e^x - e^-x)

(C) e^x + e^-x

(D) 12(e^x + e^-x)

Ques. The coefficient of x^r in the expansion of 1-ax-x²e^x is

(A) (-1)^rr!{-r² + r(a + 1) + 1}

(B) (-1)^rr!{-r² + r(a + 1) - 1}

(C) (-1)^rr!{-r² - r(a + 1) + 1}

(D) none of these

Ques. ∑_n=1^∞n²n! is equal to

(A) 2e

(B) 3e

(C) 12e

(D) none of these

Ques. The sum of the series 1 + 1+a2! + (1+a)²3! + …… is

(A) e^a - aa-1

(B) e^a - ea-1

(C) e^2a - 1a-1

(D) none of these

Ques. 1/(1·2·3) + 5/(3·4·5) + 9/(5·6·7) + ... ∞ is equal to

(A) 3/2 + 3log_e2

(B) 3/2 - 3log_e2

(C) 3/4 - 3/2 log_e2

(D) none of these

Ques. The sum of the series 2/1! + 6/2! + 12/3! + 20/4! + ... is

(A) 3e/2

(B) e

(C) 2e

(D) 3e

Ques. 5/(1·2·3) + 7/(3·4·5) + 9/(5·6·7) + ... is equal to

(A) log 8/2

(B) log_e/8

(C) log 8e

(D) none of these

Ques. [1 + 2²/2! + 2⁴/3! + 2⁶/4! + ...] / [1 + 2/2! + 2²/3! + 2⁴/4! + ...] is equal to

(A) e² + 1

(B) e² - 1

(C) e - 1

(D) e + 1

Ques. In the expansion of ln[(1+3x)/(1-2x)], the fourth term is

(A) -1/5 x³

(B) 1/4 x⁴

(C) 65/4 x⁴

(D) none of these

Ques. The coefficient of xⁿ in the expansion of log (1 + 3x + 2x²) is

(A) (-1)ⁿ/n (1 + 2ⁿ)

(B) -1/n (1 - 2ⁿ)

(C) (-1)ⁿ⁻¹/n (1 + 2ⁿ)

(D) none of these

Ques. If a = Σ(n=0 to ∞) x³ⁿ/(3n)!, b = Σ(n=0 to ∞) x³ⁿ⁻²/(3n-2)!, c = Σ(n=0 to ∞) x³ⁿ⁻¹/(3n-1)! then the value of a³ + b³ + c³ - 3abc is

(A) 1
(B) 0

(C) -1

(D) -2

Ques. If α, β are the roots of the equation x² - px + q = 0, then log_e(1 + pt + qt²) is |t| < min{1/|α|, 1/|β|}

(A) (α + β)t + (α² + β²)/2 t² + (α³ + β³)/3 t³ + ...

(B) -[(α + β)t + (α² + β²)/2 t² + (α³ + β³)/3 t³ + ...]

(C) (α + β)t - (α² + β²)/2 t² + (α³ + β³)/3 t³ - ...

(D) none of these

Ques. If n is a multiple of 3, then coefficient of xⁿ in the expansion of log_e(1 + x + x²), where -1 < x < 0, is

(A) -2/n

(B) 2/n

(C) 1/n

(D) none of these

Ques. 1·3 + (2·4)/(1·2) + (3·5)/(1·2·3) + (4·6)/(1·2·3·4) + ... ∞ is equal to

(A) 4e

(B) 5e

(C) 6e

(D) 7e

Answer to Assignement

1. B

2. A

3. A

4. B

5. A

6. B

7. D

8. A

9. B

10. C

11. C

12. A

13. C

14. A

15. A