Question

11. Evaluate (2+3*)

Answer 1

The function is ∑ k=1 11 ( 2+ 3 k )

The above function can be rewritten as,

∑ k=1 11 ( 2+ 3 k ) = ∑ k=1 11 ( 2 )+ ∑ k=1 11 ( 3 k ) =2( 11 )+ ∑ k=1 11 ( 3 k ) =22+( 3 1 + 3 2 + 3 3 ……+ 3 11 ) (1)

The terms 3 1 + 3 2 + 3 3 ……+ 3 11 are in a G.P.

Let the first term and common ratio of the given G.P. be a and r respectively.

Here,

a=3 r= 3 2 3 =3

Here, r>1

The formula for the sum of first n terms of a G.P. for r>1 is given by,

S n = a( r n −1 ) r−1

Substitute the values of a and rin equation (1) to obtain the n th term in the above expression.

S n = ∑ k=1 11 ( 3 k ) S n = 3( 3 11 −1 ) 3−1 = 3( 3 11 −1 ) 2 = 3 2 ( 3 11 −1 ) (2)

Substitute the value of equation (2) in equation (1).

∑ k=1 11 ( 2+ 3 k ) =22+ 3 2 ( 3 11 −1 )

Thus, the value of ∑ k=1 11 ( 2+ 3 k ) is 22+ 3 2 ( 3 11 −1 ).