Question

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

Answer 1

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).

S−S=[ 3+( 7−3 )+( 13−7 )+( 21−13 )+.....+( a n − a n−1 ) ]− a n 0=[ 3+( 7−3 )+( 13−7 )+( 21−13 )+.....+( a n − a n−1 ) ]− a n a n =[ 3+( 4 )+( 6 )+( 8 )+.....+( n−1 ) ] (3)

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

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

a n =3+( n−1 2 )[ 8+( n−1−1 )2 ] a n =3+( n−1 2 )[ 2n+4 ] =3+( n−1 )( 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 ).