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Question 2

2. (i) Show that

11⋅3+12⋅4+…+1N⋅(N+2)=12⁢(1+12-1N+1-1N+2),{{1}\over{1\cdot 3}}+{{1}\over{2\cdot 4}}+\dots+{{1}\over{N\cdot(N+2)}}={{1}% \over{2}}\Bigl(1+{{1}\over{2}}-{{1}\over{N+1}}-{{1}\over{N+2}}\Bigr),

where a⋅ba\cdot b denotes the product of aa and bb.

(ii) Deduce the value of the sum of the series

11⋅3+12⋅4+….{{1}\over{1\cdot 3}}+{{1}\over{2\cdot 4}}+\dots.

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