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CAT Questions | Algebra | Polynomials - Sequences

The question is about Sequences. A sequence in which terms are in fractions is given and we need to find out the sum of the sequence. 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 16: \\frac{(2^4 - 1)}{(2 - 1)}\\) + \\frac{(3^4 - 1)}{(3 - 1)}\\) + \\frac{(4^4 - 1)}{(4 - 1)}\\) + .. + \\frac{(10^4 - 1)}{(10 - 1)}\\) = ?

  1. 3462
  2. 3581
  3. 3471
  4. 4022

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Explanatory Answer

Method of solving this CAT Question from CAT Algebra - Polynomials: How do we go about simplifying this sequence?

\\frac{((2−1)(2+1)(2^2+1^2))}{(2 - 1)}\\) + \\frac{((3−1)(3+1)(3^2+1^2))}{(3 - 1)}\\) + .. + \\frac{((10−1)(10+1)(10^2+1^2))}{(10 - 1)} \\)
(2 + 1) (22 + 12) + (3 + 1) (32 + 12) + … + (10 + 1) (102 + 12)
(2 + 1) (22) + 3 + (3 + 1) (32) + 4 + (4 + 1) (42) + 5 + ..… + (10 + 1) (102) + 11
(23 + 33 + 43 + … + 103) + (22 + 32 + 42 + … + 102) + (3 + 4 + 5 + … + 11).
We know that
=> 1 + 2 + 3 + ... + n = \\frac{(n(n+1))}{2}\\)for all n > 1
=>12 + 22 + 32 + ..... + n2 = \\frac{(n(n+1)(2n+1))}{6}\\) for all n > 1

=> 13 + 23 + 33 + ..... n3 = \\frac{(n(n+1))^2}{2^2}\\)
So, in the above expression
(23 + 33 + 43 + … + 103) = 552 – 1
(22 + 32 + 42 + … + 102) = \\frac{(10 * 11 * 21)}{6}\\) - 1
(3 + 4 + 5 + … + 11) = 66 – 3 = 63
The expression simplifies as
(552 – 1) + \\frac{(10 * 11 * 21)}{6}\\) - 1 + (66 – 3) = 3471
The alternate method is to simplify \\frac{(n^4−1)}{(n−1)}\\) as n3 + n2 + n + 1 and then use the formulae for Σn3, Σn2 and Σn and Σ1

The question is "\\frac{(2^4 - 1)}{(2 - 1)}\\) + \\frac{(3^4 - 1)}{(3 - 1)}\\) + \\frac{(4^4 - 1)}{(4 - 1)}\\) + .. + \\frac{(10^4 - 1)}{(10 - 1)}\\) = ?"

Hence the answer is "3471"

Choice C is the correct answer.

 


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