Question

(1) + (1+1) + (1+1+1) +...+ (1+1+1 +...n–1 times) = ......,select a proper option (a), (b), (c) or (d) from given options so that the statement becomes correct.(All the problems refer to A.P.)
(a) (n-1)n/2
(b) n(n+1)/2
(c) n
(d) n²

Answer 1

it is given that , (1) + (1 + 1) + (1 + 1 + 1) + ........ + (1 + 1 + ...n - 1 times )
= 1 + 2 + 3 + .......... (n - 1)

here it is clear that given series is in AP , because common difference of two consecutive terms is 1 { always constant}.

first term = 1 and last term = n - 1
now, use the sum of AP formula
$S_n=\frac{n}{2}[\text{first term}+\text{last term}]$
where n is the number of terms.
here, number of term = n - 1
so, Sn = (n -1)/2 [ 1 + (n - 1)]
Sn = (n -1)/2 [ n ] = n(n -1)/2

hence option (a) is correct.

Answer 2

Formula:
The sum of first n natural number is
n(n+1)/2


1 + (1+1) + (1+1+1)........+(1+1+...n-1 terms)

= 1+2+3+....................+(n-1)

This is the sum of first (n-1) terms

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

= (n-1)n/2

= n(n-1)/2

I hope this answer helps you