The natural numbers arearranged innthe form given below

The rth group containing `2^(r-1)` numbers. Prove that sum of the numbers in the nth group is `2^(n-2)[2^(n)+2^(n-1)-1]`.

Explanation

Let 1st term of the r th goup be `T_(r)` and the 1st terms of successive rows are `1,2,4,8,"…..",` respectively. `T_(r)=1*2^(r-1)=2^(r-1)`
Hence, the sum of the numbers in te rth group is `=(2^(r-1))/(2){2*2^(r-1)+(2^(r-1)-1)*1}`
`[:. " number of terms in rth group is " 2^(r-1)]`
`=2^(r-2){2^(r)+2^(r-1)-1}`
Hence, sim of the numbers in the nth group is
`2^(r-2)[2^(n)+2^(n-1)-1]`.