Question

1^2 +3^2 +5^2 +.....+25^2=?

Answer 1

i did summation of (2n + 1)^2

Answer 2

from 0 to 25 and substituted the values...

Answer 3

shouldn't it be (2n-1)^2

Answer 4

*0 to 12

Answer 5

it could be either way i guess but if its 2n -1 we must start from 1...

Answer 6

oh you were starting from 0 then its also correct

Answer 7

sum of squares of first n odd natural numbers\[\frac{n(2n+1)(2n-1)}{3}\]

Answer 8

i was trying to derive it..

Answer 9

coz one can't remember formulas for each series..right?

Answer 10

oh sorry i thought that u were trying to solve question

Answer 11

its ok..

Answer 12

adding zero to a series wont make a difference... then the summation part should be same for both (2n+1)^2 and (2n-1)^2..but its not so..why? am i missing something?

Answer 13

\(\large \color{black}{\begin{align}
& x=1^{2} +3^{2} +5^{2} +\cdots+25^{2}\hspace{.33em}\\~\\
& x+2^{2}+4^{2}+6^{2}+\cdots+24^{2}=1^{2} +2^{2}+3^{2}+4^{2} +5^{2}+ \cdots+25^{2}\hspace{.33em}\\~\\
& x+2^{2}(1+2^{2}+3^{2}+\cdots+12^{2})=1^{2} +2^{2}+3^{2}+4^{2} +5^{2}+ \cdots+25^{2}\hspace{.33em}\\~\\
\end{align}}\)

Answer 14

sum of the squares of first natural numbers is\[= \frac{ n(n+1)(2n+1) }{ 6 }\]

substitute n=25
u can get the answer

Answer 15

@dayakar these are just odd nos. so we can't do that..

Answer 16

general term=(\(2n-1)^2\)\[T_n=4n^2-4n+1\]\[\sum4n^2-\sum4n+\sum1\]\[4 \sum n^2-4 \sum n+n\]

Answer 17

ya i sid the same way but with (2n+1)...could u do like that and show me..?

Answer 18

*did

Answer 19

if u know the sum of first n squares & n natural numbers. u can derive tht

Answer 20

ya..i do know that...but...did u read the abv part...?

Answer 21

adding zero to a series wont make a difference... then the summation part should be same for both (2n+1)^2 and (2n-1)^2..but its not so..why? am i missing something? Also ya... i did the same way but with (2n+1)...could u do like that and show me..?

Answer 22

pls help!

Answer 23

http://math.stackexchange.com/questions/1020031/find-the-exact-closed-from-expression-of-12-32-52-2n-1-2?lq=1

Answer 24

why is it (2k+1)^3 ?

Answer 25

somebody reply pls

Answer 26

\[4\sum_{?}^{?}n ^{2}-4\sum_{?}^{?}n+n\]

\[= \frac{ n(4n ^{2}-1) }{ 3 }\]

Answer 27

now plug n=25 u can get the answer

Answer 28

what were ur limits for summation?

Answer 29

in our problem n= 1 to 25

Answer 30

no..its 1 to 13 in the abv. case and anyway i know how to do this problem by this method..

Answer 31

with out using limits i find general form

Answer 32

my doubt is how to solve this:
\[\sum_{0}^{12} (2n + 1)^2\]

Answer 33

suppose if write n=3
we can get sum of squares of first 3 odd numbers

Answer 34

expand (2x+1)^2

Answer 35

\[4\sum_{0}^{12} n^2 + 4\sum_{0}^{12} n +1\]

Answer 36

sigma for 1 also

Answer 37

sigma for 1 also

Answer 38

sigma n^2 = n(n+1)(2n+1)/6


sigma n = n(n+1)/2

sig ma 1 = 12 [ limits are 1 to 12]

Answer 39

\[4[\sum_{0}^{12}\frac{ n(n+1)(2n+1) }{ 6 } ]+ 4[ \sum_{0}^{12}\frac{ n(n+1) }{ 2 }] + \sum_{0}^{12}1\]

Answer 40

sorry remove sigma

Answer 41

this summation is only for 1 to n is it the same for 0 to n also?

Answer 42

addition of 0 to any number makes any difference

Answer 43

actually i don't know clearly,

Answer 44

yes thats what i am trying to say..then why is the answer diff.?

Answer 45

what do u get

Answer 46

can u write the answer

Answer 47

in the method where we used (2n-1)^2 i m getting 2925. in this method i m getting 2924

Answer 48

i think 2925 is correct

Answer 49

ya but why r we getting to diff answers why is this method wrong?

Answer 50

in first method limits are 0 to 13

if u write n=0 and n=1 u will get both times 1
so ,we have to subtract 1 from first method answer
therefore the answer is 2924

Answer 51

is it clear

Answer 52

why are the limits 0 to 13 it must be 1 to 13 in the first method..

Answer 53

@dayakar

Answer 54

and the correct answer is...2925

Answer 55

The sum
\[ \sum_{i=0}^n 1 = n+1 \]
not n

Answer 56

yes! at last thank you!!

Answer 57

thanks again @phi .. i was confused for so long...!now i realise where i went wrong!