IIT JEE , AIEEE Forum - Mathematics , Algebra - summation of series(O-77)

neeraj_agarwal_1990

n³-(n-1)³+(n-2)³-+................(-1)^n-11³=?
If n is an odd integer >=1

[ans: (n+1)²(2n-1)/4]

I wrote the general term...
T_r+1=(n-r)³(-1)^r
now how to manipulate with that (-1)^r?

himanshu2006

PUT n= 2k+1 where k is any ineger.

now the series become:--
(2k+1)^3 -(2k)^3+ (2k-1)^3+..................... upto 'n' terms.

pair the terms with +ve sign together.i.e.:--

(2k+1)^3+(2k-1)^3+............. +1-( (2k)^3+(2k-2)^3+............+2^3)

sum the above series upto 'n'n terms & get the answer.

PLEASSSSSS RATE ME & ASK IF U UNDERGO ANY PROBLEM.

CHEERS!!!!!!!!!!!!!1

amaron

This is basically sum of odd cubes till n - sum of even cubes till n.

let n =2r - 1 implies,we need to find [1³ + 3³ ... + (2r-1)³ ] - [2³ + 4³ ... + (2r-2)³]

consider sum of cubes till n = [n(n+1)/2]² = [(2r -1)(2r)/2]²

[(2r -1)(2r)/2]²= [1³ + 3³ ... + (2r-1)³ ] + [2³ + 4³ ... + (2r-2)³]

implies

[(2r -1)(2r)/2]² - 2*[2³ + 4³ ... + (2r-2)³] = [1³ + 3³ ... + (2r-1)³ ] + [2³ + 4³ ... + (2r-2)³]

= [(2r -1)(2r)/2]²- 16*[ 1³ +2³ +3³ ...(r-1)³ ] = [(2r -1)(2r)/2]² - 16[(r(r-1)/2]² =

r² [ (2r -1)² - 4(r-1)²] as n=2r-1 , r =(n+1)/2

implies ans = (n+1)²(2n-1)/4

That solves the problem!

neeraj_agarwal_1990

I don't get it amaron

amar.gupta

Dear,

as n is an odd integer , so (n-1) will be even.

so you can write the series as :

S = 1³ - 2³+ 3³ - 4³ + ...........................-(n-1)³+n³

or S =(1³ + 2³+ 3³ + 4³ + ............+(n-1)³+n³ ) -2(2³+ 4³+6³+....(n-1)/2 terms)

{ (n-1) /2 terms as n is odd , let n=7 then it will be 3 term (2³+ 4³+6³)}

now let S1 =(2³+ 4³+6³+....(n-1)/2 terms)

or S1= 2³( 1³ + 2³+ 3³ +...........(n-1)/2 terms)

or S1= 8 [(1/2) {(n-1)/2} {(n+1)/2} ]²

{ as sum of cubes up to n tems will be [n(n+1)/2]² put n=(n-1)/2 you get the result as above}

or S1= 8 [(n-1)(n+1)/8]²

or S1= (n-1)²(n+1)²/8hence: S = [n(n+1)/2]²- 2 (n-1)²(n+1)²/8

or S = n²(n+1)²/4 - (n-1)²(n+1)²/4

or S = (n+1)² [ n² -(n-1)²] /4

or S = (n+1)² (2n-1) /4

Avinash_Bhat

Sir Amar Gupta has made it clear for you, Neeraj. I think you got him..

neeraj_agarwal_1990

got it

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