The value of the ratio (1 + 122 + 132 + 142 + ….)/(1 − 122 + 132 − 142 + ….) is …… .........
Solution
The given series are geometric series.
The first series is 1 + 12^2 + 13^2 + 14^2 + ..., where the common ratio is 12^2.
The sum of an infinite geometric series is given by S = a / (1 - r), where a is the first term and r is the common ratio.
So, the sum of the first series is S1 = 1 / (1 - 12^2).
The second series is 1 - 12^2 + 13^2 - 14^2 + ..., where the common ratio is -12^2.
So, the sum of the second series is S2 = 1 / (1 - (-12^2)).
The value of the ratio (1 + 12^2 + 13^2 + 14^2 + ...)/(1 - 12^2 + 13^2 - 14^2 + ...) is therefore S1 / S2.
However, there seems to be a mistake in the question. The common ratio in a geometric series should be the same for all terms, but here it is increasing by 1 for each term. Therefore, these are not geometric series and the formula for the sum of an infinite geometric series cannot be applied.
Please check the question and provide the correct series.
Similar Questions
The sum of first three terms of a G.P. is 1312 and their product is – 1.Find the common ratio and the terms
Find the value of 𝑟r such that the ratio of 3𝑃𝑟3 P r and 4𝑃𝑟−14 P r−1 will be 1221 ?
A and B represents the numbers of even and odd factors of 120 respectively. Find the ratio between A and B. 5:11:51:33:1
What is the quotient when 24514 divided by 113A. 217B. 216C. 215D. NONE OF THESE
The L.C.M of two numbers is 560 and their ratio is 4:5. Find the two numbers158,120130, 158102140,168112,140
Upgrade your grade with Knowee
Get personalized homework help. Review tough concepts in more detail, or go deeper into your topic by exploring other relevant questions.