deriavation of n(n+1)(2n+1)/6 i.e the sum of squares of first ‘n' natural numbers
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Hello Friend,
Here's the proof.
To find the sum of squares, we use the formula of (n+1)³ as shown in the picture.
In the proof, I have used one result directly:
1 + 2 + 3 + ... + n = n(n+1)/2
Proof:
Given series is an A.P. with
First term = 1
Last term = n
Number of terms = n
Sum = n/2 [First term + Last Term]
So, Sum = n/2 [1 + n]
So, Sum = n(n+1)/2
That is, 1 + 2 + 3 + ... + n = n(n+1)/2
I have used this result directly in the proof of sum of squares.
Hope it helps.
Purva
@Purvaparmar1405
Brainly.in
Here's the proof.
To find the sum of squares, we use the formula of (n+1)³ as shown in the picture.
In the proof, I have used one result directly:
1 + 2 + 3 + ... + n = n(n+1)/2
Proof:
Given series is an A.P. with
First term = 1
Last term = n
Number of terms = n
Sum = n/2 [First term + Last Term]
So, Sum = n/2 [1 + n]
So, Sum = n(n+1)/2
That is, 1 + 2 + 3 + ... + n = n(n+1)/2
I have used this result directly in the proof of sum of squares.
Hope it helps.
Purva
@Purvaparmar1405
Brainly.in
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I added a new image. I have written more clearly there
u gave only one page photo right??so i want the simplification of the where it is written (n+1)^3 -1^3=3S +3n(n+1)/2+n
we need S to get the answer so...can u show the simplifications after the step
At that place, take LCM in denominator. Here, LCM is 2. So all terms on RHS are multiplied by 2
and what about (n+1)^3 -1^3...??
There I used the identity:
(n+1)³ = n³ + 3n² + 3n + 1
This 1 gets cancelled with (-1) already present
i could not understand properly but anyways thanks for u help support
help n support*
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