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The reason why I have splited 4 = 2+2 ; 7 = 4+3 ; 11 = 5+7, is since the difference between the consecutive terms are in AP, if I subtract that difference from the particular term of the sequence I will get the previous term. This way I can split the whole sequence into an AP sequence and the given sequence with n-1 terms.

Re: Help:Tough problem on exponential sequence [#permalink]
24 Jul 2010, 20:18

I've learnt this method at school, and it can be applied easily to a wide variety of problems tht involve summations.

On observation we find that the 'n'th term of this sequence is of the form

T_n=T_{n-1}+ [n-1]

for n >= 2 i.e from the 2nd term onwards.

so to find the 60th term (T_60) all we need to do is this :

T_{60} = T_{59} + (60-1) = T_{59}+59 T_{59} = T_{58}+58 and so on . . . . T_3 = T_2+2 T_2 = T_1+1 ___________

On summing all the equations' left hand sides and right hand sides, observe that every equations' LHS (starting with the second) cancels with the term in the RHS above it - for example T59 in the second eqn in the LHS cancels with T59 in the RHS of the first equation and so on for the entire system of equations.

After this huge crash of dominoes, all we are left with is

T_{60} = T_1 + (1+2+3+4+...59)

1+2+...59 as we know is {59*(59+1)}/2 = 1770 and T_1 = 1

Re: Help:Tough problem on exponential sequence [#permalink]
28 Jan 2011, 11:16

2

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Expert's post

144144 wrote:

I might missing something small... how u done 1+2+3+4+5+6+7+8+9+10....+59+60 in a short way? can someone please explain the logic?

is there a difference if the amount of numbers is an even or uneven number?

thanks a lot.

First of all we have 1+(1+2+3+...+59).

Next, the sum of the elements in any evenly spaced set is given by: Sum=\frac{first+last}{2}*# \ of \ terms, the mean multiplied by the number of terms. So, 1+(1+2+3+...+59)=1+\frac{1+59}{2}*59=1+1770=1771.

Re: Help:Tough problem on exponential sequence [#permalink]
28 Jan 2011, 11:41

Bunuel wrote:

144144 wrote:

I might missing something small... how u done 1+2+3+4+5+6+7+8+9+10....+59+60 in a short way? can someone please explain the logic?

is there a difference if the amount of numbers is an even or uneven number?

thanks a lot.

First of all we have 1+(1+2+3+...+59).

Next, the sum of the elements in any evenly spaced set is given by: Sum=\frac{first+last}{2}*# \ of \ terms, the mean multiplied by the number of terms. So, 1+(1+2+3+...+59)=1+\frac{1+59}{2}*59=1+1770=1771.