Question

The sum of first n terms of the series 1^2+2.2^2+3^2+2.4^2.........is n(n+1)^2, when n is even. When n is odd

Answer 1

the sum is

Answer 2

@Queen_Bee1 this is the question

Answer 3

ohhh ok

Answer 4

well i got 20.6

Answer 5

no but we have to represent the answer in terms of n

Answer 6

@wio help

Answer 7

What exactly is the sequence?

Answer 8

\[1^{2}+2.2^{2} + 3^{2}+2.4^{2}.......\]

Answer 9

2.2?

Answer 10

What is the sequence in terms of \(n\)?

Answer 11

sum is n(n+1)^2/2 when n is even we have to fnd when n is odd

Answer 12

Ok, if \(n\) is odd, then it is just \(S_{n+1} - a_n\)

Answer 13

ok

Answer 14

And \(S_{n+1}\) is even, so you can use your formula.

Answer 15

When \(n\) is odd, it seems \(a_n\) is \(n^2\).

Answer 16

Does that help you?

Answer 17

can u pls write the equation

Answer 18

You want a direct answer?

Answer 19

yes

Answer 20

Okay, plug \(n+1\) into the answer for the even sum. Then subtract \(n^2\). That is the answer.

Answer 21

pls can u write direct answer

Answer 22

I'm too lazy.

Answer 23

pls

Answer 24

\[
\frac{(n+1)((n+1)+1)^2}2 - n^2\]

Answer 25

but the answer is \[n^2(n+1)/2\]

Answer 26

wait a moment, is that the answer for even or odd?

Answer 27

for odd

Answer 28

hmmm, well mine needs to be simplified.

Answer 29

that or the even formula was wrong?

Answer 30

is that giving the same

Answer 31

oh wait. I think I should have done \(-a_{n+1}\)

Answer 32

so \(-2n^2\)

Answer 33

Eh, I didn't put a lot of thought into it.

Answer 34

@ganeshie8 PLS HELP

Answer 35

i think wio's method is using triangular numbers, and after representing the nth term you need to find the sum...

Answer 36

My method is finding the sum for \(n+1\), since we know the even sum, and then just subtracting the last \(n+1\) term, to get the \(n\) sum.

Answer 37

Oh okay... looks i misinterpreted the solution

Answer 38

\[
S_n = S_{n+1}-a_{n+1}
\]

Answer 39

If \(n\) is odd then\(n+1\) is even. We know how to do the even sum. That was my logic.

Answer 40

wow! yes last term is tricky, got you :)

Answer 41

got u @wio

Answer 42

this question Is a bit tricky and I have a test ON 6TH

Answer 43

can any 1 write the solution

Answer 44

pls

Answer 45

@dan815 @hartnn @ganeshie8 @ParthKohli @batmann

Answer 46

did u get wio's logic ?

Answer 47

yes

Answer 48

is this the series?\[1^2 + (2\times 2^2) + 3^2 + (2\times 4^2)\cdots\]

Answer 49

no I have written thee series above

Answer 50

You're given a formula which works when \(n\) is even :
\[\large 1^2+2.2^2+3^2+2.4^2 + \cdots + 2.n^2= n(n+1)^2 \]

Answer 51

yes

Answer 52

its

Answer 53

and you're trying to find a formula that works when \(n\) is odd :

\[\large 1^2+2.2^2+3^2+2.4^2 + \cdots + 2.(n-1)^2 + n^2 = ? \]

Answer 54

write it like this
(1^2+2^2+3^2+...+n^2) + (2^2+4^2+6^2+...n^2)
Now, try to find the sum.

Answer 55

yes

Answer 56

\[= (1^2 + 2^2 + 3^2 + \cdots n^2) + (2^2 + 4^2 + 6^2\cdots n
^2)\]

Answer 57

we're already given the formula for sum when "n" is even

Answer 58

yeah

Answer 59

we just need to use it and deduce a formula for odd "n"

Answer 60

ok

Answer 61

which was exactly what @wio did above ^

Answer 62

thnx @ganeshie8 @wio

Answer 63

if you have it like this
\(\sum_{n=1}^n n(n+1)^2 \)
why you need to find odd /even ?

Answer 64

let me rewrite the wio's solution again in one reply

Answer 65

or this is what your qn asking for ?
how to find sum when n is odd and find the sum when n is even ?

Answer 66

\[\large \color{red}{1^2+2.2^2+3^2+2.4^2 + \cdots + 2.(n-1)^2} + n^2 \]

\[\large \color{red}{\dfrac{(n-1)(n-1+1)^2}{2}}+ n^2 \]

\[\large \color{red}{\dfrac{(n-1)(n)^2}{2}}+ n^2 \]

\[\large \color{red}{\dfrac{(n-1)(n)^2 + 2n^2}{2}}\]

\[\large \color{red}{\dfrac{n^2(n+1)}{2}}\]

Answer 67

ikram : we're given a formula for `even n`, our job is to deduce the formula for `odd n`

Answer 68

thnx for u support every1

Answer 69

oh , ic ^^
for once get confused

Answer 70

lol same wid me, i went straight for the proof before wio hinting me

Answer 71

huh

Answer 72

ok

Answer 73

should I close the problem