1.

Consider the sequence in the form of group (1),(2,2)(3,3,3),(4,4,4,4),(5,5,5,5,5…..) The sum of first 2000 terms isA. 84336B. 96324C. 78466D. none of these

Answer» Correct Answer - A
1,2,2,3,3,3,4,4,4,4,..
Let us write the terms in the groups as follows: 1,(2,2),(3,3,3),(4,4,4,4),… consisting of 1,2,3,4,.. Terms. Let 2000th term fall in nth group. Then,
`((n-1)n)/2lt2000le(n(n+1))/2`
or n(n-1)`lt4000len(n+1)`
Let us consider,
`n(n-1)lt4000`
or `n^(2)-n-4000lt0`
or `nlt(1+sqrt(16001))/2` r `nlt64`
We have
`n(n+1)ge4000` or `n^(2)+n-4000ge0` or `nge63`
That means 2000th term falls in 63rd group, which means that the 2000th term is 63. Now, the total number of terms up to 62nd group is `(62xx63)//2=1953`. Hence, the sum of first 2000 terms is
`1^(2)+2^(2)+..+62^(2)+63(2000-1953)`
`=(62(63)125)/6+63xx47=84336`
Sum of the remaining terms=`63xx16=1008`.


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