Question

# $$\displaystyle { 1 }^{ 2 }+({ 1 }^{ 2 }+{ 2 }^{ 2 })+({ 1 }^{ 2 }+{ 2 }^{ 2 }+{ 3 }^{ 2 })+....$$

Solution

## $$\displaystyle \text{The given series is}, \quad { 1 }^{ 2 }+({ 1 }^{ 2 }+{ 2 }^{ 2 })+({ 1 }^{ 2 }+{ 2 }^{ 2 }+{ 3 }^{ 2 })+...\\ { a }_{ n }=({ 1 }^{ 2 }+{ 2 }^{ 2 }+{ 3 }^{ 2 }+...+{ n }^{ 2 })\\ =\dfrac { n(n+1)(2n+1) }{ 6 } \\ =\dfrac { n({ 2n }^{ 2 }+3n+1) }{ 6 } =\dfrac { { 2 }^{ 3 }+{ 3n }^{ 2 }+n }{ 6 } \\ =\dfrac { 1 }{ 3 } { n }^{ 3 }+\dfrac { 1 }{ 2 } { n }^{ 2 }+\dfrac { 1 }{ 6 } { n }\\ \therefore { S }_{ n }=\sum _{ k=1 }^{ n }{ { a }_{ k } } \\ =\sum _{ k=1 }^{ n }{ \left( { \dfrac { 1 }{ 3 } { k }^{ 2 }+\dfrac { 1 }{ 2 } { k }^{ 2 }+\dfrac { 1 }{ 6 } { k } } \right) } \\ =\dfrac { 1 }{ 3 } \sum _{ k=1 }^{ n }{ { k }^{ 3 } } +\dfrac { 1 }{ 2 } \sum _{ k=1 }^{ n }{ { k }^{ 2 } } +\frac { 1 }{ 36 } \sum _{ k=1 }^{ n }{ { k } } \\ =\dfrac { 1 }{ 3 } \dfrac { { n }^{ 2 }(n+1)^{ 2 } }{ { (2 })^{ 2 } } +\dfrac { 1 }{ 2 } \times \dfrac { n(n+1)(2n+1) }{ 6 } +\dfrac { 1 }{ 6 } \times \dfrac { n(n+1) }{ 2 } \\ =\dfrac { n(n+1) }{ 6 } \left[ \dfrac { n(n+1) }{ 2 } +\dfrac { (2n+1) }{ 2 } +\dfrac { 1 }{ 2 } \right] \\ =\dfrac { n(n+1) }{ 6 } \left[ \dfrac { { n }^{ 2 }+n+2n+1+1 }{ 2 } \right] \\ =\dfrac { n(n+1) }{ 6 } \left[ \dfrac { { n }^{ 2 }+n+2n+2 }{ 2 } \right] \\ =\dfrac { n(n+1) }{ 6 } \left[ \dfrac { { n(n+1)+2(n+1) } }{ 2 } \right] \\ =\dfrac { n(n+1) }{ 6 } \left[ \dfrac { (n+1)(n+2) }{ 2 } \right] \\ =\dfrac { n{ (n+1) }^{ 2 }(n+2) }{ 12 }$$MathematicsNCERTStandard XI

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