Q15 of 85 Page 110

(1) + (1 + 1) + (1 + 1 + 1) + ... + (1 + 1 + 1 + ...n — 1 times) =

(1) + (1 + 1) + (1 + 1 + 1) + .…. + (1 + 1 + 1 + ...n — 1 times)


= 1 + 2 + 3 + 4 ………… + (n – 1)


= Sn – 1


⇒ Sn – 1 = × [2a + (n – 1 – 1)d]


= × [2(1) + (n – 2)(1)]


=


=


=


=


∴ the correct option is (a).

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