Question

(4-1/n)+(4-2/n)+(4-3/n)+..... N

Answer 1

Answer:

Let the given series be x

x = (4-1/n) + (4-2/n) + (4-3/n) + .....

  = ( 4+4+4....) - (1/n+2/n+3/n+....)

  = x₁-x₂

• x₁ = 4( 1+1+1...)

a=1, d=0

x₁ = 4*n/2[ 2*1+( n-1)*0] ( since xₙ= n/2[ 2a+( n-1)*d]

⇒ x₁  = 4n

• x₂ = 1/n(1+2+3 ....)

a = 1, d = 2-1=1

x₂ = 1/n*n/2 [2*1+(n-1)*1]

    = 1/2 (2+n-1)

⇒ x₂ = 1/2 ( 1+n)

Thus x= x₁-x₂

x= 4n-1/2 ( 1+n )

x= (8n-1-n)/2

x= 7n-1/ 2

HENCE THE SUM OF N TERM SERIES= 7n-1/2

Answer 2

Answer:

we have to find:-

 the sum of 4-1/n, 4-2/n, 4-3/n = ?

solution:-

we know that :

1 + 2 +3 + 4 +5 +6 + ...........+ n = n(n +1) / 2

and 

1 + 1+ 1 + 1 + 1 + .............+ n = n 

Here,

sum of 4-1/n, 4-2/n, 4-3/n up to the nth term 

= (4 + 4 + 4 + 4 + 4 + ......... upto n terms) + (-1/n - 2/n - 3/n - ..........upto n terms)

= 4 ( 1+1+1+1.......... upto n terms) - 1/n (1 + 2 + 3 +4 .........upto n terms)

= 4 n - 1/n × n(n +1)/2

= 4n - (n+1)/2 

=  [ 8n - (n+1) ] / 2     .......taking L.C.M 

=( 7n - 1) / 2 Answer