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Question

Let  $$S _ { k } = \dfrac { 1 + 2 + 3 + \ldots + k } { k } .$$  If   $${ S }_{ { 1 } }^{ { 2 } }+{ S }_{ { 2 } }^{ { 2 } }++{ S }_{ { 10 } }^{ { 2 } }=\dfrac { 5 }{ 12 } { A }$$  then  $$A$$  is equal to :


A
303
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B
283
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C
156
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D
301
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Solution

The correct option is A $$303$$
$$\mathrm { S } _ { \mathrm { K } } = \dfrac { \mathrm { K } + 1 } { 2 }$$
$$\Sigma S _ { k } ^ { 2 } = \dfrac { 5 } { 12 } A$$
$$\sum _ { \mathrm { k } = 1 } ^ { 10 } \left( \dfrac { \mathrm { K } + 1 } { 2 } \right) ^ { 2 } = \dfrac { 2 ^ { 2 } + 3 ^ { 2 } + \dots  + 11 ^ { 2 } } { 4 } = \dfrac { 5 } { 12 } \mathrm { A }$$
$$\dfrac { 11 \times 12 \times 23 } { 6 } - 1 = \dfrac { 5 } { 3 } \mathrm { A }$$
$$505 = \dfrac { 5 } { 3 } \mathrm { A } , \quad \mathrm { A } = 303$$

Mathematics

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