Sum to $n$ terms of the series $\dfrac { 1 }{ 1.2.3.4 } +\dfrac { 1 }{ 2.3.4.5 } +\dfrac { 1 }{ 3.4.5.6 } +..........$, is
- $\dfrac { { n }^{ 3 } }{ 3\left( n+1 \right) \left( n+2 \right) \left( n+3 \right) }$
- $\dfrac { { n }^{ 3 }+{ 6n }^{ 2 }-3n }{ 6\left( n+2 \right) \left( n+3 \right) \left( n+4 \right) }$
- $\dfrac { 15{ n }^{ 2 }+7n }{ 4n\left( n+1 \right) \left( n+5 \right) }$
- $\dfrac { { n }^{ 3 }+6{ n }^{ 2 }+11n }{ 18\left( n+1 \right) \left( n+2 \right) \left( n+3 \right) }$
Reveal answer
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D
Correct answer
Explanation
The general term of the series is 1/(n(n+1)(n+2)(n+3)). Using the method of differences, this can be written as (1/3) * [1/(n(n+1)(n+2)) - 1/((n+1)(n+2)(n+3))]. Summing this telescoping series leads to the expression in option D.
AI explanation
The general term is a_n = 1/[n(n+1)(n+2)(n+3)]. Using partial fractions, this term becomes 1/3[1/(n(n+1)(n+2)) - 1/((n+1)(n+2)(n+3))], creating a telescoping series. Summing this from 1 to n gives 1/3[1/(1*2*3) - 1/((n+1)(n+2)(n+3))], which simplifies to 1/18 - 1/[3(n+1)(n+2)(n+3)]. Combining these over a common denominator results in (n^3 + 6n^2 + 11n)/[18(n+1)(n+2)(n+3)].