(1),(2,3),(4,5,6),(7,8,9,10),................the sum of terms in 50th bracket=?62,525 6

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Mirka (2910)

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Consider sum of first terms of each group

S = 1 + 2 + 4 + 7 + 11 + …. T_n

S = 1 + 2 + 4 + 7 + ……T_n-1 + T_n

Subtract them,

0 = 1 + ( 1 + 2 + 3 + … n-1 ) - T_n

T_n = 1 + n(n-1)/2 = (n² –n + 2) / 2 …(1)

Now, nth group contains ‘n’ terms

So, 50^th gp contains 50 terms, all terms in A.P wid common diff = 1

And first term of 50^th gp = 1226 from formula 1

So sum = (use A.P formula ) = 50/2 { 1226*2 + 1*(50-1) }

= 62525

-na-

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