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If S=41⋅2⋅3+52⋅3⋅4+63⋅4⋅5⋯S = \dfrac{4}{1 \cdot 2 \cdot 3} + \dfrac{5}{2 \cdot 3 \cdot 4} + \dfrac{6}{3 \cdot 4 \cdot 5} \cdots S=1⋅2⋅34+2⋅3⋅45+3⋅4⋅56⋯ till nnn terms and S=K−(2n+5)2(n+1)(n+2)S = K - \dfrac{(2n+5)}{2(n+1)(n+2)} S=K−2(n+1)(n+2)(2n+5), then find K KK.
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