IIT JEE , AIEEE Forum - Mathematics , Algebra

shailesh_45

1) 1 + [ 1/ 1+2] + [ 1 / 1+ 2 + 3 ] .........n

2) 1/[ 1 + 1^2 +1^4 ] + 2/ [1 + 2^2 + 2^4] + 1 / [1+ 3^2 + 3^4 ] ......n

3 ) 1/ [1-x][1-x^3] + x^2/[1-x^3][1-x^5] + x^4/[1-x^5][1-x^7]

4) s + a +b + c + d prove [s-a][s-b][s-c][s-d] > 81abcd

5) bc , ca ,ab are in AP a> 0,b>0 ,c>0

prove a^n + c^n >2b^n

hash_include

1.
$\sum_{i=1}^{n} \frac 1 {\frac {r(r+1)}{2}}$

$= 2. \sum_{i=1}^{n} \frac 1{r(r+1)}$

$= 2. \sum_{i=1}^{n} \frac {(r+1) - r }{r(r+1)}$

$= 2. [\sum_{i=1}^{n} \frac 1 {r} - \sum_{i=1}^{n} \frac 1 {r+1}]$

$= 2[ (\frac 1 1 - \frac 1 2) + (\frac 1 2 - \frac 1 3) ....... + (\frac 1 n - \frac 1 {n+1})]$

everything except the first and last term gets cancelled

$= 2 ( 1 - \frac 1 {n+1})$

$= \frac {2n}{n+1}$

Smile

computer001

2nd write denom as:
r^4 + r^2 +1 = (r^2+r+1)(r^2-r+1)
write num as r^2+r+1 -{r^2-r+1}
then devide into 2 terms u will have telescopic cancellation as in the prev sum

shailesh_45

complete soln required

computer001

wth???
c mate..im not postin here to earn salutes..im postin 2 help u...
so dont demand..i think if u read thru my soln u can urself do the sum...learn to help urself

hash_include

dont be so rude computer001 :D

computer001

wth??
im speakin perfectly fine...some ppl need to understand tht
not all ppl post for the sake of salutes

computer001

4) s=a+b+c+d
s-a=b+c+d
s-b=a+c+d ....

so [s-a][s-b][s-c][s-d]=(b+c+d)(a+b+c)(a+c+d)+(a+b+d)
for each of the 4 terms apply AM-GM

hash_include

theres a typo in that qsn..
is the ans comin for s = a + b + c + d??

computer001

last q what is n?
i mean naturals or sumtin can u b more specific?

shailesh_45

don't be angry please vvvvvvvsorry but please do not use any short forms for anything i cannot unserdtand them

shailesh_45

n = natural nos

hash_include

last qsn i'm gettin a,b,c are in HP.. after that im not sure what to do..

computer001

{a^n + c^n}/2 >=
{a+c}^n/2^n >=
[4/{1/a+1/c}]^n/2^n
=
[2b]^n/2^n
=b^n
as a,b,c in H.P

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