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Algebra

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20 May 2009 14:17:03 IST

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

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(1),(2,3),(4,5,6),(7,8,9,10),................the sum of terms in 50th bracket=?62,525 62,255 65,225

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Sujit

Blazing goIITian

Joined: 11 Mar 2009

Posts: 1345

20 May 2009 20:23:31 IST

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ast term of Nth bracket is given by n(n+1)/2

so the last term of 49th brack is 1225

therefore the require sum is 25[2*1226 +49*1] = 62525

Mirka

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Joined: 13 Aug 2008

Posts: 1313

21 May 2009 19:39:59 IST

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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

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