Given, Sn=131+13+231+2+13+23+331+2+3+…+13+23+…+n31+2+…+n
Tn=(2n(n+1))22n(n+1)
Tn=n(n+1)2=2(n1−n+11)
Sn=2n=1∑n(n1−n+11)
Sn=2(1−n+11)=n+12n
Given, 100Sn=n⇒100×n+12n=n
n+1=200
n=199
Let Sn=131+13+231+2+13+23+331+2+3+…+13+23+…n31+2+…,+n . If 100Sn=n, then n is equal to:
Held on 9 Apr 2017 · Verified 6 Jul 2026.
200
199
99
19
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