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Question

Sum of all the products of the first n positive integers taken two at a time is?


A
124(n1)n(n+1)(3n+2)
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B
148(n2)n(n1)n2
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C
16(n+1)(n+2)(n+5)
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D
None of these
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Solution

The correct option is A $$\displaystyle\frac{1}{24}(n-1)n(n+1)(3n+2)$$
$$\left( { x }_{ 1 }+{ x }_{ 2 }+{ x }_{ 3 }+......{ x }_{ n } \right) ^{ 2 }$$
$$=\left( { x }_{ 1 }^{ 2 }+{ x }_{ 2 }^{ 2 }+{ x }_{ 3 }^{ 2 }+......{ x }_{ n }^{ 2 } \right) +2$$(Sum of product of numbers taken two at a time)
$$\Rightarrow 2\times $$ (Sum of product of numbers taken two at a time)
$$=\left[ \dfrac { n(n+1) }{ 2 }  \right] ^{ 2 }-\cfrac { n(n+1)(2n+1) }{ 6 } $$
$$\Rightarrow $$ (Sum of product of numbers taken two at a time)$$\cfrac { 1 }{ 4 } n(n+1)\left[\dfrac{ n(n+1) }{ 2 } -\cfrac { 2n+1 }{ 3 }  \right] $$
$$= \cfrac { 1 }{ 24 } n(n+1)(n-1)(3n+2)$$
Hence the correct answer is $$\cfrac { 1 }{ 24 } n(n+1)(n-1)(3n+2)$$

Mathematics

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