Question

Find the sum of n terms of the series $(4-\frac{1}{n})+(4-\frac{2}{n})+(4-\frac{3}{n})+...$

Answer 1

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