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Question

The sum of (n + 1) terms of 11+11+2+11+2+3+...... is
[RPET 1999]

A
nn+1
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B
2nn+1
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C
2n(n+1)
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D
2(n+1)n+2
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Solution

The correct option is D 2(n+1)n+2
Tn=1[n(n+1)2]=2[1n1n+1]
Put n = 1, 2, 3, ...., (n+1)
T1=2[1112],T2=2[1213],.....,
Tn+1=2[1n+11n+2]
Hence sum of (n + 1) terms =n+1k+1
Sn+1=2[11n+2]Sn+1=2(n+1)(n+2).

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