IIT JEE , AIEEE Forum - Mathematics , Algebra

sidhart16

${1}^{3} - {2}^{3} + {3}^{3} - {4}^{3} +...+{9}^{3}$ =

Ans is given 425

please give me a simple detailed method of how to solve it easily rates assured

a) 425 b) -425 c) 475 d) -475

LAMPARD

Most sensible thing is to add the cubes directly...that would save much more time than any other method...

jessforyou

the qn can be simplified to sum(a^3-(a+1)^3) i.e

$\sum_{1}^{8}(-({x}^{2}+x+1))$ + ${9}^{3}$

which can be found using the formulae

sidhart16

i cld not understands lampad plz give me other easy method that method given by u not applicable if the series is very big

LAMPARD

OK..see...

The series can be written as follows

$\sum_{1}^{5}$ $(2r-1)^{3}$ - $(2r)^{3}$

Which comes out to be

$\sum_{1}^{5}$ -4 ${r}^{2}$ +2r -1

Now,its just series expansion...sry for taking a long time...using the new editor for the first time..

studyid

lampard bhai ......

ur method is correct .......but seems as if u have forgot to put the summation sign over the second term . ..... i.e.

(2r)³

LAMPARD

No no..the summation sign includes both..infact i forgot to add a term,i.e, +10^3 as -10^3 comes out as a term if u expand the summation which we dont want...

studyid

ya perfect .....soln ....... sorry .......

sidhart16

$\sum_{1}^{5}$ $(2r-1)^{3}$ - $(2r)^{3}$

@ Lampard

is there any trick to get the above form or we will have to get them by thinking

sidhart16

-sorry lamard got got odd 2n - 1

even 2n

thanks

pantpranav

The easiest way to solve it is to directly put the values because we are dealing with small numbers.

It would be a wastage of time using any formula.

hsbhatt

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