Determine the second smallest prime factor of 13-11-1-23-12-1-33-13-1- -20053-12005-1- correct answer - 3- wrong answer 0

Let S=1^3+11+1+2^3+12+1+3^3+13+1+ ⋯+2005^3+12005+1 ⇒ S=∑_r=1^2005(r^2 r+1) ⇒ S=16(2005 × 2006 ×4011) nbsp;12(2005 × 2006)+2005 ⇒ S=5 × 401 × 1340009 ⇒ S=5 × ...

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Standard XII

Mathematics

Sigma n2

Determine the...

Question

Determine the second smallest prime factor of $\frac{1^{3} + 1}{1 + 1} + \frac{2^{3} + 1}{2 + 1} + \frac{3^{3} + 1}{3 + 1} + \dots + \frac{2005^{3} + 1}{2005 + 1} .$
(correct answer + 3, wrong answer 0)

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Solution

Let $S = \frac{1^{3} + 1}{1 + 1} + \frac{2^{3} + 1}{2 + 1} + \frac{3^{3} + 1}{3 + 1} + \dots + \frac{2005^{3} + 1}{2005 + 1}$
$\Rightarrow S = 2005 \sum r = 1 (r^{2} - r + 1) \Rightarrow S = \frac{1}{6} (2005 \times 2006 \times 4011) - \frac{1}{2} (2005 \times 2006) + 2005 \Rightarrow S = 5 \times 401 \times 1340009 \Rightarrow S = 5 \times 11 \times 401 \times 121819$

Since $2, 3, 7$ does not divide $401$ and $121819,$ the second smallest prime factor is $11.$

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Determine the second smallest prime factor of 13+11+1+23+12+1+33+13+1+⋯+20053+12005+1. (correct answer + 3, wrong answer 0)

Let S=13+11+1+23+12+1+33+13+1+⋯+20053+12005+1 ⇒S=2005∑r=1(r2−r+1)⇒S=16(2005×2006×4011)−12(2005×2006)+2005⇒S=5×401×1340009⇒S=5×11×401×121819 Since 2,3,7 does not divide 401 and 121819, the second smallest prime factor is 11.

Determine the second smallest prime factor of $\frac{1^{3} + 1}{1 + 1} + \frac{2^{3} + 1}{2 + 1} + \frac{3^{3} + 1}{3 + 1} + \dots + \frac{2005^{3} + 1}{2005 + 1} .$
(correct answer + 3, wrong answer 0)