Question

Finf the sum of (4-1/n)+(4-2/n)+(4-3/n) +....... upto n terms

Answer 1

It's easy..

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

⇒ 4n- (1/n+2/n+3/n...n/n)

⇒ 4n-[(1+2+3...n)/n]

⇒ 4n-[n(n+1)/2/n] [∵ 1+2+3..n = n(n+1)/2]

⇒4n-(n+1)/2

⇒(8n-n-1)/2

⇒(7n-1)/2

That's it !