11)3+7+14+27+52....
t1=0+1*3
t2=1+2*3
t3=2+4*3
t4=3+8*3
tn=(n-1)+2(n-1)*3
therefore sum= summation (n-1)+ sumation(3*2n-1)
FIND THE SUM upto n terms of the following series:-------
1) 3+7+14+27+52.......
2) (1)+(2+3)+(4+5+6)+......
3) (33-23)+(53-43)+(73-63)+..........
4) 1+22-33+4+52-63+7+82-93.......
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4 Answers
1
12) last term of nth bracket is (n+1)n/2
therefore summation of the series is summation of 1st (n+1)n/2 natural nos
3) this is of the form {(2n+1)3-2n3}
(2n+1-2n){(2n+1)2+(2n+1)2n+(2n)2}
(12n2+6n+1)
summation of (12n2+6n+1)
4n3+9n2+6n ans
14) of the form {(3n-2)+(3n-1)2-(3n)3}
summation of this
(3n-2)(3n-1)(13n-9n2-2)/4
7@Chintan:
IN Q2 --- if i find the 1st term of d nth bracket which comes up to be = 1+{n(n-1)/2} and then find the sum of d terms of the nth bracket.then if i want to find the sum of (n-1) brackets then what shud i do?
though it may be a lendy approach still then.....
In q4) How come u make up this form? by just observing or any work-out????