1. N the set of natural numbers is partitioned into subsets $$S_{1}$$ = $$(1)$$, $$S_{2}$$ = $$(2,3)$$, $$S_{3}$$ =$$(4,5,6)$$, $$S_{4}$$ = $${7,8,9,10}$$ and so on. The sum of the elements of the subset $$S_{50}$$ is
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By: anil on 05 May 2019 02.34 am
According to given question $$S_{50}$$ will have 50 terms
And its first term will be 50th number in the series 1,2,4,7,.........$$T_{50}$$
$$T_1 = 1$$
$$T_2 = 1+1$$
$$T_3 = 1+1+2$$
$$T_4 = 1+1+2+3$$
$$T_n = 1+(1+2+3+4+5....(n-1))$$
= $$1+frac{n(n-1)}{2}$$
So $$T_{50} = 1+1225 = 1226$$
Hence $$S_{50} = (1226,1227,1228,1229........)$$
And summation will be = $$frac{50}{2} (2 imes 1226 + 49 imes 1 ) = 62525$$
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And its first term will be 50th number in the series 1,2,4,7,.........$$T_{50}$$
$$T_1 = 1$$
$$T_2 = 1+1$$
$$T_3 = 1+1+2$$
$$T_4 = 1+1+2+3$$
$$T_n = 1+(1+2+3+4+5....(n-1))$$
= $$1+frac{n(n-1)}{2}$$
So $$T_{50} = 1+1225 = 1226$$
Hence $$S_{50} = (1226,1227,1228,1229........)$$
And summation will be = $$frac{50}{2} (2 imes 1226 + 49 imes 1 ) = 62525$$