We have,
1+12+141+1+22+242+1+32+343+…
Here,
tn=1+n2+n4n
=(n2+n+1)(n2−n+1)n
So,
t1=21(1−31)
t2=21(31−71)
⋮⋮⋮⋮⋮⋮⋮⋮⋮
t10=21(911−1111)
Now,
S10=t1+t2+t3+.....+t10
⇒S10=21(1−1111)
⇒S10=11155
The sum to 10terms of the series
1+12+141+1+22+242+1+32+343+…is :-
Held on 1 Feb 2023 · Verified 6 Jul 2026.
11159
11155
11156
11158
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