Question

Find $\frac{1}{2}+\frac{1^{2}+2^{2}}{3}+\frac{1^{2}+2^{2}+3^{2}}{4}+\frac{1^{2}+2^{2}+3^{2}+4^{2}}{4}+....$ upto n terms.

Answer 1

Solution :

$\frac{1}{2}+\frac{1^{2}+2^{2}}{3}+\frac{1^{2}+2^{2}+3^{2}}{4}+\frac{1^{2}+2^{2}+3^{2}+4^{2}}{4}+....$ upto n terms.

nth term in the series ( tn )

= [ 1²+2²+3²+ ...+n² ]/( n + 1 )

= [ n( n+1 )( 2n + 1 ) ]/[ 6( n + 1 ) ]

= [ n( 2n + 1 ) ]/6 ----- ( 1 )

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We know that ,

i ) Sum of first n natural Numbers

= n(n+1)/2

ii ) Sum of the squares of first n

natural numbers =[ n(n+1)(2n+1)/6 ]

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Sum of the terms = Sn

= Sigma [ n(2n+1)/6 ]

= 1/6 { Sigma ( 2n² + n ) }

= 1/6 { 2sigma n² + sigma n }

= 1/6 { 2[ n(n+1)(2n+1)/6 ] + [ n(n+1)/2 ] }

= 1/6 { [n(n+1)(2n+1)/3] + [n(n+1)/2 ] }

= 1/6 { n(n+1)[ (2n+1)/3 + 1/2 ] }

= [n(n+1)/6] { [ 2(2n+1) + 3 ]/6 }

= [ n(n+1)/6] { (4n + 2 + 3 )/6]

Sn = [ n(n+1)( 4n+5 )]/36

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