Answer to Question #195060 in Discrete Mathematics for Maxwell

Question #195060

Using mathematical induction prove the following:

Σn k=1  k+4/k(k + 1)(k+2) = n(3n + 7)/2(n+1)(n+2) 
1
Expert's answer
2021-05-19T12:19:44-0400

Check for n = 1:

k=11k+4k(k+1)(k+2)=1+41(1+1)(1+2)=56=1(31+7)2(1+1)(1+2)\sum\limits_{k = 1}^1 {\frac{{k + 4}}{{k(k + 1)(k + 2)}} = \frac{{1 + 4}}{{1(1 + 1)(1 + 2)}} = \frac{5}{6}} = \frac{{1\left( {3 \cdot 1 + 7} \right)}}{{2(1 + 1)(1 + 2)}} - equality holds.

Let the equality hold for n = s, i. e

k=1sk+4k(k+1)(k+2)=s(3s+7)2(s+1)(s+2)\sum\limits_{k = 1}^s {\frac{{k + 4}}{{k(k + 1)(k + 2)}} = } \frac{{s(3s + 7)}}{{2(s + 1)(s + 2)}}

Let us check the equality for n = s+1:

k=1s+1k+4k(k+1)(k+2)=s(3s+7)2(s+1)(s+2)+s+1+4(s+1)(s+2)(s+3)=s(3s+7)(s+3)+2(s+5)2(s+1)(s+2)(s+3)=3s3+16s2+21s+2s+102(s+1)(s+2)(s+3)=3s3+6s2+3s+10s2+20s+102(s+1)(s+2)(s+3)=3s(s2+2s+1)+10(s2+2s+1)2(s+1)(s+2)(s+3)=(3s+10)(s+1)22(s+1)(s+2)(s+3)=(3s+10)(s+1)2(s+2)(s+3)=(s+1)(3(s+1)+7)2((s+1)+1)((s+1)+2)\sum\limits_{k = 1}^{s + 1} {\frac{{k + 4}}{{k(k + 1)(k + 2)}} = } \frac{{s(3s + 7)}}{{2(s + 1)(s + 2)}} + \frac{{s + 1 + 4}}{{(s + 1)(s + 2)(s + 3)}} = \frac{{s(3s + 7)(s + 3) + 2(s + 5)}}{{2(s + 1)(s + 2)(s + 3)}} = \frac{{3{s^3} + 16{s^2} + 21s + 2s + 10}}{{2(s + 1)(s + 2)(s + 3)}} = \frac{{3{s^3} + 6{s^2} + 3s + 10{s^2} + 20s + 10}}{{2(s + 1)(s + 2)(s + 3)}} = \frac{{3s({s^2} + 2s + 1) + 10({s^2} + 2s + 1)}}{{2(s + 1)(s + 2)(s + 3)}} = \frac{{(3s + 10){{(s + 1)}^2}}}{{2(s + 1)(s + 2)(s + 3)}} = \frac{{(3s + 10)(s + 1)}}{{2(s + 2)(s + 3)}} = \frac{{(s + 1)(3(s + 1) + 7)}}{{2((s + 1) + 1)((s + 1) + 2)}}

So, equality holds for n=s+1. Then equality holds for all nNn \in N , Q.E.D.


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