CAT Quantitative Aptitude Questions | CAT Algebra - Polynomials questions

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

Video Explanation

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.