help to get an answer

manisha

New member
Joined
Oct 24, 2012
Messages
11
Hello friends ,


I trying to solve some question papers where i came across this question

if (1^2+2^2+3^2....+10^2)=385 then the value of (2^2+4^2+6^2....+20^2) is:

can anbody help me to get an answer ............

or tricks to get the answer.

in shortcut way .......or any idea

thanku.................​
 
Hello friends ,


I trying to solve some question papers where i came across this question

if (1^2+2^2+3^2....+10^2)=385 then the value of (2^2+4^2+6^2....+20^2) is:

can anbody help me to get an answer ............

or tricks to get the answer.

in shortcut way .......or any idea

thanku.................​
Note \(\displaystyle 2^2 = (2 * 1)^2 = 2^2 * 1^2 = 4 * 1^2.\)

\(\displaystyle 4^2 = (2 * 2)^2 = 2^2 * 2^2 = 4 * 2^2.\)

\(\displaystyle 6^2 = (2 * 3)^2 = 2^2 * 3^2 = 4 * 3^2.\)

And so on. Can you do the problem now?
 
Hello friends ,


I trying to solve some question papers where i came across this question

if (1^2+2^2+3^2....+10^2)=385 then the value of (2^2+4^2+6^2....+20^2) is:

can anbody help me to get an answer ............

or tricks to get the answer.

in shortcut way .......or any idea

thanku.................​

Notice that the first set of numbers are all in the form of some number squared, or n^2. The second set of numbers takes those same numbers and doubles them (or multiplies them all by 2), so they can be represented by (2n)^2. We can simplify (2n)^2 to be (2^2)(n^2). Does that help?
 
Notice that the first set of numbers are all in the form of some number squared, or n^2. The second set of numbers takes those same numbers and doubles them (or multiplies them all by 2), so they can be represented by (2n)^2. We can simplify (2n)^2 to be (2^2)(n^2). Does that help?

Sorry

i tried in this way and got the answer 1600 which is wrong ,

the correct answer is 2485

I dont know how it is?
 
Sorry

i tried in this way and got the answer 1600 which is wrong ,

the correct answer is 2485

I dont know how it is?

1600 is not correct, nor is 2485.

Please double check both your problem statement and the given answer.

Please show your work when doing calculations.
 
Hello friends ,


I trying to solve some question papers where i came across this question

if (1^2+2^2+3^2....+10^2)=385​


There's your first problem, this sum is 285, not 385.
so 2^2+ 4^2+... 20^2= 4(1+ 2^2+ 3^2+ ...+ 10^2)= 4(285)= 1140.

then the value of (2^2+4^2+6^2....+20^2) is:

can anbody help me to get an answer ............

or tricks to get the answer.

in shortcut way .......or any idea

thanku.................
 
Hello, manisha!

\(\displaystyle \text{If }\,1^2 + 2^2 + 3^2 + \cdots + 10^2 \,=\,385.\)

\(\displaystyle \text{find the value of: }\:2^2 + 4^2 + 6^2 + \cdots + 20^2\)

We are told: .\(\displaystyle \displaystyle\sum^{10}_{k=1} k^2 \:=\:385\)
We are asked to find: .\(\displaystyle \displaystyle\sum^{10}_{k=1} (2k)^2 \)

Hence: .\(\displaystyle \displaystyle \sum^{10}_{k=1}(2k)^2 \;=\;\sum^{10}_{k=1}4k^2 \;=\;4\sum^{10}_{k=1}k^2 \;=\;4\cdot 385 \;=\;1540\)
 
Makes no sense to give this: if (1^2+2^2+3^2....+10^2)=385
PROBABLY meant the odds only:
if (1^2 + 3^2 + 5^2.....+17^2 + 19^2) = 1330

Using sum of squares formula, 1 to 20 (n=20):
n(n+1)(2n+1)/6 = 20(21)(41)/6 = 2870

Since 2870 includes the odds: 2870 - 1330 = 1540

hey...
how do u got 1330
as, using the formula ,where n=19

19(19+1)(2*19+1)/6=14820/6 =2470

how?


thanks.....................
 
hey...
how do u got 1330
as, using the formula ,where n=19

19(19+1)(2*19+1)/6=14820/6 =2470

how?


thanks.....................
WJM and I gave you a hint sufficient to solve this problem as you posed it. Denis suggested an alternative approach. Soroban actually worked out the problem as you posed it along the lines of the hint given by wjm and me.

Here is the situation: (a) you posed the problem incorrectly, (b) you misunderstood the answer given in the book, or (c) the book's answer is a typographical error. The question you posed has been answered now multiple times, and you yourself can check that that answer is correct with a calculator.
 
Top