one of the best

Algebra Level 5

Let,

$\large S_n = a_1^n + a_2^n + a_3^n + a_4^n + a_5^n + a_6^n + a_7^n + a_8^n + a_9^n + a_{10}^n + a_{11}^n + a_{12}^n$

Given that,

$\large S_1 = S_2 = S_3 = S_4 = S_6 = S_8 = S_9 = S_{11} = 0$

$\large S_5 = 35$ , $\large S_7 = 14$ , $\large S_{10} = 245$ , $\large S_{12} = 168$

Find the largest value of $n$ for which $\large S_n = 0$

This problem is a part of the sets - 3's & 4's & QuEsTiOnS

The answer is 23.

1 solution

Avinash Singh
Mar 23, 2015

using newton sum we get

p1=s1=0 p2=0=p3=p4=0 p5=7 p6=0 p7=2 p8=0 p9=0 p10=0 p11=0 p12=0

now again using newton sum to find S13

$S13=S12P1-S11P2+S10P3-S9P4+S8P5-S7P6+S6P7-S5P8+S4P9-S3P10+S2P11-S1P12$

IF WE USE THE ABOVE RESULT THE EQATION BECOMES

$S13=S6P7+S8P5$ NOW PUTTING THE VALUE OF P7 AND P5 WE GET $S13=2S6+7S8$ look at this it can be written as n for every integer as $S_{n+7}=2S_{n}+7S_{n+2}$ SINCE WE NEED TO FIND HIGHEST VALUE OF N+7 SO S_{n+7}=0 NOW EQATING WITH ZERO WE GET

$-2S_{n}=7S_{n+2}$ NOW THIS IS ENOUGH FOR THE RESULT....LOOK AT THIS

$-2S_{9}=7S_{11}..............BECAUSE BOTH S_{9} AND S_{11} =0$

BY THIS WE GET S16=0.........SIMILARLY DOING WITH S11 AND S13 BOTH ARE EQUAL TO ZERO. WE GET..............S18=0

NOW WITH S16 AND S18 WHICH ARE EQUAL TO ZERO AND HAVE DIFFERENCE 2 i.e(18-16=2)............WE GET S23=0

NOW THERE ARE NO SUCH PAIR VALUE OF S WHICH ARE EQAL TO ZERO AND DIFFER BY 2.

SO $S_{23}=0$

You should prove why $S_n > S_{23}$ will not be equal to zero

Sakanksha Deo - 6 years, 2 months ago

BECAUSE...........23-2=21..........AND S21 IS NOT EQAL TO ZERO ......IF IT BE SO THERE WOULD BE THE NEXT VALUE FOR S{n}

Avinash Singh - 6 years, 2 months ago

one of the best

This problem is a part of the sets - 3's & 4's & QuEsTiOnS

The answer is 23.

1 solution

0 pending reports