Resourcesdown chevron
Testimonials
Plans
Sign in
Sign up
Resourcesright arrow
Testimonials
Plans
Bytelearn - cat image with glassesAI tutor

Welcome to Bytelearn!

Let’s check out your problem:

If the number of terms in (x+1+(1)/(x))^(n)(n inI^(+))is 401 , then n is greater than (a) 201 (b) 200 (c) 199 (d) None of these

88. If the number of terms in (x+1+1x)n(n∈I+) \left(x+1+\frac{1}{x}\right)^{n}\left(n \in I^{+}\right) is 401401 , then n n is greater than\newline(a) 201201\newline(b) 200200\newline(c) 199199\newline(d) None of these

View step-by-step helpView step-by-step help
Homeright chevron icon
Math Problemsright chevron icon
Algebra 2right chevron icon
Sum of finite series not start from 1right chevron icon

Full solution

Q. 88. If the number of terms in (x+1+1x)n(n∈I+) \left(x+1+\frac{1}{x}\right)^{n}\left(n \in I^{+}\right) is 401401 , then n n is greater than\newline(a) 201201\newline(b) 200200\newline(c) 199199\newline(d) None of these
  1. Expansion of Expression: The number of terms in the expansion of (x+1+1x)n(x+1+\frac{1}{x})^n is given by n+1n+1 because the expression can be rewritten as (x+(1+1x))n(x + (1 + \frac{1}{x}))^n, which is a binomial expression with two terms.
  2. Find nn: To find nn, we set n+1n+1 equal to 401401.\newlinen+1=401n + 1 = 401
  3. Solve for n: Subtract 11 from both sides to solve for n.\newlinen=401−1n = 401 - 1\newlinen=400n = 400
  4. Compare nn to Options: Now we compare nn to the options given.n=400n = 400 is greater than 199199 and 200200 but not greater than 201201.

More problems from Sum of finite series not start from 1

Question
Find the sum of the finite arithmetic series. ∑n=110(7n+4)\sum_{n=1}^{10} (7n+4)\newline______
Get tutor helpright-arrow

Posted 2 months ago

Question
Type the missing number in this sequence: `55,\59,\63,\text{_____},\71,\75,\79`
Get tutor helpright-arrow

Posted 1 month ago

Question
Type the missing number in this sequence: `2, 4 8`, ______,`32`
Get tutor helpright-arrow

Posted 2 months ago

Question
What kind of sequence is this? 2,10,50,250,…2, 10, 50, 250, \ldots Choices:Choices:\newline[A]arithmetic\text{[A]arithmetic}\newline[B]geometric\text{[B]geometric}\newline[C]both\text{[C]both}\newline[D]neither\text{[D]neither}
Get tutor helpright-arrow

Posted 2 months ago

Question
What is the missing number in this pattern? 1,4,9,16,25,36,49,64,81,____1, 4, 9, 16, 25, 36, 49, 64, 81, \_\_\_\_
Get tutor helpright-arrow

Posted 1 month ago

Question
Classify the series. ∑n=012(n+2)3 \sum_{n = 0}^{12} (n + 2)^3 \newlineChoices:\newline[A]arithmetic\text{[A]arithmetic}\newline[B]geometric\text{[B]geometric}\newline[C]both\text{[C]both}\newline[D]neither\text{[D]neither}
Get tutor helpright-arrow

Posted 1 month ago

Question
Find the first three partial sums of the series.\newline1+6+11+16+21+26+⋯1 + 6 + 11 + 16 + 21 + 26 + \cdots\newlineWrite your answers as integers or fractions in simplest form.\newlineS1=S_1 = ____\newlineS2=S_2 = ____\newlineS3=S_3 = ____
Get tutor helpright-arrow

Posted 1 month ago

Question
Find the third partial sum of the series. \newline3+9+15+21+27+33+⋯3 + 9 + 15 + 21 + 27 + 33 + \cdots\newlineWrite your answer as an integer or a fraction in simplest form. \newlineS3=S_3 = ____
Get tutor helpright-arrow

Posted 1 month ago

Question
Find the first three partial sums of the series. \newline1+7+13+19+25+31+⋯1 + 7 + 13 + 19 + 25 + 31 + \cdots\newlineWrite your answers as integers or fractions in simplest form. \newlineS1=S_1 = ____\newlineS2=S_2 = ____\newlineS3=S_3 = ____
Get tutor helpright-arrow

Posted 2 months ago

Question
Does the infinite geometric series converge or diverge?\newline1+34+916+2764+⋯1 + \frac{3}{4} + \frac{9}{16} + \frac{27}{64} + \cdots\newlineChoices:\newline[A]converge\text{[A]converge}\newline[B]diverge\text{[B]diverge}
Get tutor helpright-arrow

Posted 1 month ago

View all (97)