Forgetting factorials in e x (every term after the first two has n! in the denominator). Mixing the signs in ln(1 + x) (it alternates: +, −, +, −, …). Using the log series when |x| ≥ 1 (it won’t be valid or will be very slow). Confusing bases: ln means base e ; log (in many school texts) often means base 10.

Forgetting factorials in e x (every term after the first two has n! in the denominator). Mixing the signs in ln(1 + x) (it alternates: +, −, +, −, …). Using the log series when |x| ≥ 1 (it won’t be valid or will be very slow). Confusing bases: ln means base e ; log (in many school texts) often means base 10.

Exponential series: Safe to use for any input; it behaves well in general. Logarithm series: Best when the number you are taking the log of is close to one. If the number is far from one, this method becomes slow or unhelpful.

Exponential series: Safe to use for any input; it behaves well in general. Logarithm series: Best when the number you are taking the log of is close to one. If the number is far from one, this method becomes slow or unhelpful.

Yes. They measure the same idea on different scales. You can move from one to the other by multiplying or dividing by a fixed conversion factor. A common approach is to estimate using the natural version and then convert to the base-ten version if your book requires it.

Yes. They measure the same idea on different scales. You can move from one to the other by multiplying or dividing by a fixed conversion factor. A common approach is to estimate using the natural version and then convert to the base-ten version if your book requires it.

CBSE Class 12 Maths Notes on Exponential and Logarithmic Series

Q: It is a step-by-step way to build the value of the exponential function. You start with a basic value and keep adding smaller and smaller corrections. Because each correction is tiny compared with the previous one, the total settles to the true value.

It is a step-by-step way to build the value of the exponential function. You start with a basic value and keep adding smaller and smaller corrections. Because each correction is tiny compared with the previous one, the total settles to the true value.

Q: Take only the first few corrections and stop when the next correction would be very small. For the logarithm series (near one), the corrections usually switch between adding and removing. For the exponential series, the corrections are all additions but get tiny very quickly. In most classroom problems, a few corrections give a good answer.

Take only the first few corrections and stop when the next correction would be very small. For the logarithm series (near one), the corrections usually switch between adding and removing. For the exponential series, the corrections are all additions but get tiny very quickly. In most classroom problems, a few corrections give a good answer.

Exponential Series

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

Some Particular Cases

e = 1 + 11! + 12! + 13! + …… ∞
e^-1 = 1 - 11! + 12! - 13! + …… ∞
e^-x = 1 - x1! + x²2! - x³3! + …… to ∞
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

(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!

(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

(D) e - 3

Ques. ∑_n=2^∞n²n! is equal to

(A) 2e

(B) 3e

(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

(D) e/4

Logarithmic Series

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

Some Particular Cases

Replacing x by -x, |x| < 1: log_e(1 - x) = -x - x²2 - x³3 - x⁴4 - …… to ∞
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

(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

(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)

(D) none of these

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

A) x + 1x

(B) 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)

(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

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

(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

(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

(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

(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)!

(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

(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)

(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)

(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

(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

(D) 524

Sol. (A). (1 - 3x + x²) e^-x

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

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

Frequently Asked Questions

Forgetting factorials in e^x (every term after the first two has n! in the denominator).
Mixing the signs in ln(1 + x) (it alternates: +, −, +, −, …).
Using the log series when |x| ≥ 1 (it won’t be valid or will be very slow).
Confusing bases: ln means base e; log (in many school texts) often means base 10.

It is a step-by-step way to build the value of the exponential function. You start with a basic value and keep adding smaller and smaller corrections. Because each correction is tiny compared with the previous one, the total settles to the true value.

Exponential series: Safe to use for any input; it behaves well in general.
Logarithm series: Best when the number you are taking the log of is close to one. If the number is far from one, this method becomes slow or unhelpful.

Take only the first few corrections and stop when the next correction would be very small. For the logarithm series (near one), the corrections usually switch between adding and removing. For the exponential series, the corrections are all additions but get tiny very quickly. In most classroom problems, a few corrections give a good answer.

S.no	Formulas List
1.	Trigonometric equations
2.	Solution of the triangle
3.	Height and distance
4.	Inverse trigonometric function
5.	Exponential and logarithm series
6.	Logarithm & properties
7.	Quadratic equations
8.	Indefinite integral
9.	Definite integrals
10.	Differential equation

Exponential and logarithm series

Exponential Series

Some Particular Cases

Logarithmic Series

Some Particular Cases

Formulae and Concepts at a Glance

Frequently Asked Questions

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Frequently Asked Questions