CAT Quantitative Aptitude Questions | CAT Algebra - Polynomials questions

CAT Questions | Algebra | Polynomials - Sum of a Series

The question is about Sum of a Series. The same series is given in two forms and we need to find out the sum of the series. Polynomials is a simple topic which involves a lot of basic ideas. Make sure that you a get of hold of them by solving these questions. A range of CAT questions can be asked based on this simple concept. -->

Question 18: What is the sum of \\frac{7}{1}\\) + \\frac{26}{2}\\) + \\frac{63}{3}\\) + \\frac{124}{4}\\) + \\frac{215}{5}\\) .... 19 terms or 7 + 13 + 21 + 31 + 43 + 57 + 73... 19 terms?

  1. 3100
  2. 3025
  3. 3044
  4. 3097

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Explanatory Answer

Method of solving this CAT Question from CAT Algebra - Polynomials: How you would go about simplifying the series?

\\frac{(2^3−1)}{(2−1)}\\) + \\frac{(3^3−1)}{(3−1)}\\) + \\frac{(4^3−1)}{(4−1)}\\) + \\frac{(5^3−1)}{(5−1)}\\) +\\frac{(6^3−1)}{(6−1)}\\) …
\\frac{(a^3−1)}{(a−1)}\\) = a2 + a + 1
So, the above expression becomes
(22 + 2 + 1) + (32 + 3 + 1) + (42 + 4 + 1) + ………. + (202 + 20 + 1)
Σn2 + Σn + Σ1 from 2 to 20
\\frac{(n(n+1)(2n+1))}{6}\\) + \\frac{(n(n+1)}{2}\\) + n-3 (for n = 20)
n * (\\frac{(n(n+1)(2n+1))}{6}\\) + \\frac{(n+1)}{2}\\) + 1) - 3 (for n = 20)
n * (\\frac{(2n^2+6n+10)}{6}\\)) - 3
n * (\\frac{(20(400+60+5))}{3}\\)) - 3 = 3100 - 3 = 3097

The question is "What is the sum of \\frac{7}{1}\\) + \\frac{26}{2}\\) + \\frac{63}{3}\\) + \\frac{124}{4}\\) + \\frac{215}{5}\\).... 19 terms or 7 + 13 + 21 + 31 + 43 + 57 + 73... 19 terms?"

Hence the answer is "3097"

Choice D is the correct answer.

 


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