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Calculus Examples
∞∑n=1(23)n∞∑n=1(23)n
Step 1
The sum of an infinite geometric series can be found using the formula a1-ra1−r where aa is the first term and rr is the ratio between successive terms.
Step 2
Step 2.1
Substitute anan and an+1an+1 into the formula for rr.
r=(23)n+1(23)nr=(23)n+1(23)n
Step 2.2
Cancel the common factor of (23)n+1(23)n+1 and (23)n(23)n.
Step 2.2.1
Factor (23)n(23)n out of (23)n+1(23)n+1.
r=(23)n23(23)nr=(23)n23(23)n
Step 2.2.2
Cancel the common factors.
Step 2.2.2.1
Multiply by 11.
r=(23)n23(23)n⋅1r=(23)n23(23)n⋅1
Step 2.2.2.2
Cancel the common factor.
r=(23)n23(23)n⋅1
Step 2.2.2.3
Rewrite the expression.
r=231
Step 2.2.2.4
Divide 23 by 1.
r=23
r=23
r=23
r=23
Step 3
Since |r|<1, the series converges.
Step 4
Step 4.1
Substitute 1 for n into (23)n.
a=(23)1
Step 4.2
Simplify.
a=23
a=23
Step 5
Substitute the values of the ratio and first term into the sum formula.
231-23
Step 6
Step 6.1
Multiply the numerator by the reciprocal of the denominator.
23⋅11-23
Step 6.2
Simplify the denominator.
Step 6.2.1
Write 1 as a fraction with a common denominator.
23⋅133-23
Step 6.2.2
Combine the numerators over the common denominator.
23⋅13-23
Step 6.2.3
Subtract 2 from 3.
23⋅113
23⋅113
Step 6.3
Multiply the numerator by the reciprocal of the denominator.
23(1⋅3)
Step 6.4
Cancel the common factor of 3.
Step 6.4.1
Factor 3 out of 1⋅3.
23(3⋅1)
Step 6.4.2
Cancel the common factor.
23(3⋅1)
Step 6.4.3
Rewrite the expression.
2
2
2