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Question

23.Find the sum of the first n terms of the series: 3+ 7 +13 +21 +31 +

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Solution

The given series is 3+7+13+21+31+.....

S=3+7+13+21+31+.....(1)

S=0+3+7+13+21+.....(2)

Subtract equation (2) from equation (1).

SS=[ 3+( 73 )+( 137 )+( 2113 )+.....+( a n a n1 ) ] a n 0=[ 3+( 73 )+( 137 )+( 2113 )+.....+( a n a n1 ) ] a n a n =[ 3+( 4 )+( 6 )+( 8 )+.....+( n1 ) ] (3)

It can be seen that equation (3) represents an AP.

So, nth term of an AP can be formulated as,

a n =3+( n1 2 )[ 8+( n11 )2 ] a n =3+( n1 2 )[ 2n+4 ] =3+( n1 )( n+2 ) = n 2 +n+1

Sum of the series can be represented as,

k=1 n k 2 = k=1 n k 2 + k=1 n k + k=1 n 1 = n( n+1 )( 2n+1 ) 6 + n( n+1 ) 2 +1 =n[ ( n+1 )( 2n+1 )+3( n+1 )+6 6 ] =n[ ( n+1 )( 2n+1 )+3( n+1 )+6 6 ]

Further simplify the equations.

k=1 n k 2 = k=1 n k 2 + k=1 n k + k=1 n 1 =n[ 2 n 2 +6n+10 6 ] = n 3 ( n 2 +3n+5 )

Thus, sum of the series 3+7+13+21+31+..... is n 3 ( n 2 +3n+5 ).


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