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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]`. |
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Answer» 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]`. |
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