Question
The sum of (n+1) terms of 11+11+2+11+2+3+.......... is
Answer: Option D
:
D
Tn=1[n(n+1)2]=2[1n−1n+1]
put n=1,2,3,.........,(n+1)
T1=2[11−12],T2=2[12−13],...........,
Tn+1=2[1n+1−1n+2]
Hence sum of (n+1) terms = n+1∑k=1Tk
⇒Sn+1=2[1−1n+2] ⇒Sn+1=2(n+1)n+2
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:
D
Tn=1[n(n+1)2]=2[1n−1n+1]
put n=1,2,3,.........,(n+1)
T1=2[11−12],T2=2[12−13],...........,
Tn+1=2[1n+1−1n+2]
Hence sum of (n+1) terms = n+1∑k=1Tk
⇒Sn+1=2[1−1n+2] ⇒Sn+1=2(n+1)n+2
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