
Concept explainers
State the proof of the given statement.

Explanation of Solution
Given:
The given statement is 11⋅3+12⋅4+13⋅5+⋅⋅⋅+1n(n+2)=n(3n+5)4(n+1)(n+2) for all natural numbers.
Calculation:
Use mathematical induction.
S1=n(3n+5)4(n+1)(n+2)=1(3⋅1+5)4(1+1)(1+2)=13
Sk=11⋅3+12⋅4+13⋅5+⋅⋅⋅+1k(k+2)=k(3k+5)4(k+1)(k+2)
Sk+1=11⋅3+12⋅4+13⋅5+⋅⋅⋅+1k(k+2)+1(k+1)(k+3)=(k+1){3(k+1)+5}4{(k+1)+1}{(k+1)+2}=(k+1){3k+8}4{k+2}{k+3}
Show that Sk+1(k+1)(k+3)=(k+1)(3k+8)4(k+2)(k+3)
From L.H.S
Sk+1(k+1)(k+3)
[Sk=k(3k+5)4(k+1)(k+2)]
=k(3k+5)4(k+1)(k+2)+1(k+1)(k+3)
=k(3k+5)(k+3)+4(k+2)4(k+1)(k+2)(k+3)
=(3k2+5k)(k+3)+4(k+2)4(k+1)(k+2)(k+3)
=3k3+5k2+9k2+15k+4k+84(k+1)(k+2)(k+3)
=3k3+3k2+11k2+11k+8k+84(k+1)(k+2)(k+3)
=3k2(k+1)+11k(k+1)+8(k+1)4(k+1)(k+2)(k+3)
=(k+1)(3k2+11k+8)4(k+1)(k+2)(k+3)
=3k2+11k+84(k+2)(k+3)
=3k2+3k+8k+84(k+2)(k+3)
=3k(k+1)+8(k+1)4(k+2)(k+3)
=(k+1)(3k+8)4(k+2)(k+3)=RH.S
Hence the given statement is true for all natural numbers.
Chapter EP Solutions
Algebra 2
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