For a sequence t_1, t_2, t_3, ..., S_n denotes the sum of the first n

hey there, how did you establish logical connexion between m and n and how did you conclude that \(t_m = S_m – S_{m-1} \)

[quote="Bunuel"]For a sequence \(t_1\), \(t_2\), \(t_3\), ..., \(S_n\) denotes the sum of the first n terms of a sequence. If \(S_n = n^3 + n^2 + n + 1\), and \(t_m = 291\), then m is equal to?


S1 = 4; S2 = 15; S3 = 40; S4=85

So t1, t2, t3 .... = 4, 11, 25, 45...

=> difference of the terms t1, t2, t3 is in AP series

=> Sn = 4 + 11 + 25 + 45 + .... tm .............................(1)
and Sn =.......4 + 11 + 25 + ..... (m-1) terms + tm .........................(2)

Subtract (2) from (1)

=> 0 = 4 + 7 + 14 + 20 + ...... to m-1 terms - tm
=> tm = 4 + (7+14+20+.....to m-1 terms)
=> tm = 4 + (2*7+ (m-1-1)*7)(m-1) /2
=> tm = 4+ (14+(m-2)*7)(m-1)/2

if we replace m=1 we get t1 = 4; m=2 => t2 =11; m3 => t3 = 25.. hence our eqn is correct.

so, for tm = 291; similarly we can replace values of m from options to get LHS = RHS. But none of the options are matching..Even for m=10, we get tm = 319 and not 291; There is some error in the question IMO

How to solve these questions on a GMAT Whiteboard. I got the concept of using Sm - S(m-1), but putting values made it tough to calculate in whiteboard. And as it was taking more than 2 minutes, I went for Option B as it ended as 1.
22*22*22 - 21*21*21 + 22*22 - 21*21 + 22 -21 = 1431
1000 - 729 + 100 - 81 + 10-9 = 291

hi there allan89 can you pls elaborate on highlighted part Dont get what we are doing there

difference is not constant i.e. 7 and 6 (not equal) . can you please elaborate it?

both two equations are same. in the second equation it is written as one term shifting to get the difference constant (although imo there is not constant difference).

s1 = sum of first term = t1
s2= sum of the first two terms = t1+t2
s3= sum of the first three terms = t1+t2+t3
s4= sum of the first four terms = t1+t2+t3+t4

so, s2-s1 =( t1+t2) - t1 = t2
s3-s2 =( t1+t2+t3) - (t1+t2) = t3

hope, it is clear now.

Bunuel GMATWhizTeam
S1=1^3+1^2+1+1=4 ;t1= 3*1^2-1+1 =3 how is it possible?
for this reason, s2 and (t1+t2) , s3 and (t1+t2+t3) all are making difference by 1. where am I wrong? please help.

If I look at the question from GMAT perspective and solve it within the gambit of what is asked in GMAT, then I will look for a pattern.
Now
\(S_n = n^3 + n^2 + n + 1\)
1) When n=1, \(S_n = n^3 + n^2 + n + 1\) => \(S_1 = 1^3 + 1^2 + 1 + 1=4\)....\(S_1=t_1=4\)
2) When n=2, \(S_n = n^3 + n^2 + n + 1\) => \(S_2 = 2^3 + 2^2 + 2 + 1=15\)....\(S_2=t_1+t_2=4+t_2=15....t_2=11\)
3) When n=3, \(S_n = n^3 + n^2 + n + 1\) => \(S_3 = 3^3 + 3^2 + 3 + 1=40\)....\(S_3=S_2+t_3=15+t_3=40....t_3=25\)
4) Similarly \(t_4=45........ t_5=71........ t_6=103\)
Sequence is 4, 11, 25, 45, 71, 103.....
We can search either for next 3-4 values as 291 is not a very big value to look for OR find a pattern in the sequence..
So, let us look for the sequence..
The sequence is 4, 11, 25, 45, 71, 103.....
Difference in consecutive numbers is 7, 14, 20, 26, 32, That if after first term, the difference increases by 6..
Thus the sequence becomes 4, 11, 25, 45, 71, 103, 103+(32+6)=141, (141)+(38+6)=185, 185+(44+6)=235, 235+(50+6)=291(our term)
so sequence is 4, 11, 25, 45, 71, 103, 141, 185, 235, 291, and 291 is the \(10^{th}\) term.

A­

Hi chetan2u

could you pls explain what does this formula mean and what do we find by \(T_n\)? is it like recursive or what ?

\(T_n = S_n – S_{n-1} \)

thanks ­

For a sequence \(t_1\), \(t_2\), \(t_3\), ..., \(S_n\) denotes the sum of the first n terms of a sequence. If \(S_n = n^3 + n^2 + n + 1\), and \(t_m = 291\), then m is equal to?

A. 10--> correct:

S_m= m^3 + m^2 + m + 1=(m+1)(m^2+1) => t_m = S_m-S_{m-1}= (m+1)(m^2+1) - ((m-1)+1)((m-1)^2+1)= (m+1)(m^2+1) - m(m^2-2m+1+1)= (m+1)(m^2+1) - m(m^2-2m+2) = m^3 + m^2 + m + 1 - m^3+2m^2-2m=3m^2-m+1=291=>3m^2-m-290=0=>3m^2-30m+29m-290=0=>(m-10)(3m+29)=0, m>=0, so,m=10

B. 22
C. 24
D. 26
E. 30

Hey, I would like to pitch in here.
consider a series- 1,2,3,4,5,6
Sum of first 5 no.s- 5(5+1)/2 = 15
Sum of first 6 no.s- 6(6+1)/2 = 21

Difference between S6 and S5 = 21-15 = 6 -> which is same as 6th term­

I think what people got wrong about this question is to treat the series as arithmetic series. Hence, there are many people who chose (B) as an answer.

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