# Polynomials

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## Sum of a Series

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

Choice D. 3097

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## Detailed Solution

${\frac{(2^3−1)}{(2−1)} + \frac{(3^3−1)}{(3−1)} + \frac{(4^3−1)}{(4−1)} + \frac{(5^3−1)}{(5−1)} + \frac{(6^3−1)}{(6−1)}}$
${\frac{(a^3−1)}{(a−1)} = a^2 + a + 1}$
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)
$n * (\frac{( 2n^2+6n+10)}{6}) - 3$
$n * (\frac{( 20(400+60+5))}{3}) - 3 = 3100 - 3 = 3097$

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