The sum of (n + 1) terms of 11+11+2+11+2+3+...... is [RPET 1999]
A
nn+1
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
2nn+1
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
2n(n+1)
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
2(n+1)n+2
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
Open in App
Solution
The correct option is D2(n+1)n+2 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+1 ⇒Sn+1=2[1−1n+2]⇒Sn+1=2(n+1)(n+2).