# Polynomials

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## Sequences

${\frac{(2^4 - 1)}{(2 - 1)} + \frac{(3^4 - 1)}{(3 - 1)} + \frac{(4^4 - 1)}{(4 - 1)} + .. + \frac{(10^4 - 1)}{(10 - 1)} = ?}$
1. 3462
2. 3581
3. 3471
4. 4022

Choice C. 3471

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## Detailed Solution

${\frac{((2−1)(2+1)(2^2+1^2))}{(2 - 1)} + \frac{((3−1)(3+1)(3^2+1^2))}{(3 - 1)} + .. + \frac{((10−1)(10+1)(10^2+1^2))}{(10 - 1)} }$

$(2 + 1) (2^2 + 1^2) + (3 + 1) (3^2 + 1^2) + … + (10 + 1) (10^2 + 1^2)$

$(2 + 1) (2^2) + 3 + (3 + 1) (3^2) + 4 + (4 + 1) (4^2) + 5 + ..… + (10 + 1) (10^2) + 11$

$(2^3 + 3^3 + 4^3 + … + 10^3) + (2^2 + 3^2 + 4^2 + … + 10^2) + (3 + 4 + 5 + … + 11).$

We know that

-> ${1 + 2+ 3 + ... + n = \frac{(n(n+1))}{2}}$ for all n > 1

->${1^2 + 2^2 + 3^2 + ..... + n^2 = \frac{(n(n+1)(2n+1))}{6} }$ for all n > 1

-> ${1^3 + 2^3 + 3^3 + ..... n^3 = (\frac{(n(n+1))}{2})^2 }$

So, in the above expression

${(2^3 + 3^3 + 4^3 + … + 10^3) = 55^2 – 1}$

${(2^2 + 3^2 + 4^2 + … + 10^2) = \frac{(10 x 11 x 21)}{6} - 1 }$

(3 + 4 + 5 + … + 11) = 66 – 3 = 63

The expression simplifies as

${(55^2 – 1) + \frac{(10 x 11 x 21)}{6} - 1 + (66 – 3) = 3471 }$

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

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