Question

The sum of squares of first n natural numbers is given by n(n+1)(2n+1)/6 . Find the sum of squares of the first 10 natural numbers.​

Answer 1

$</p><p>\huge{\underline{\underline{\underline{\sf{\pink{ﾑɳรᏇɛƦ \: ࿐}}}}}}</p><p>$

We will discuss here how to find the sum of the squares of first n natural numbers.

We will discuss here how to find the sum of the squares of first n natural numbers.Let us assume the required sum = S

We will discuss here how to find the sum of the squares of first n natural numbers.Let us assume the required sum = STherefore, S = 122 + 222 + 322 + 422 + 522 + ................... + n22

We will discuss here how to find the sum of the squares of first n natural numbers.Let us assume the required sum = STherefore, S = 122 + 222 + 322 + 422 + 522 + ................... + n22Now, we will use the below identity to find the value of S:

We will discuss here how to find the sum of the squares of first n natural numbers.Let us assume the required sum = STherefore, S = 122 + 222 + 322 + 422 + 522 + ................... + n22Now, we will use the below identity to find the value of S:n33 - (n - 1)33 = 3n22 - 3n + 1

We will discuss here how to find the sum of the squares of first n natural numbers.Let us assume the required sum = STherefore, S = 122 + 222 + 322 + 422 + 522 + ................... + n22Now, we will use the below identity to find the value of S:n33 - (n - 1)33 = 3n22 - 3n + 1Substituting, n = 1, 2, 3, 4, 5, ............., n in the above identity, we get

We will discuss here how to find the sum of the squares of first n natural numbers.Let us assume the required sum = STherefore, S = 122 + 222 + 322 + 422 + 522 + ................... + n22Now, we will use the below identity to find the value of S:n33 - (n - 1)33 = 3n22 - 3n + 1Substituting, n = 1, 2, 3, 4, 5, ............., n in the above identity, we get                     133 - 033 = 3 . 122 - 3 ∙ 1 + 1

We will discuss here how to find the sum of the squares of first n natural numbers.Let us assume the required sum = STherefore, S = 122 + 222 + 322 + 422 + 522 + ................... + n22Now, we will use the below identity to find the value of S:n33 - (n - 1)33 = 3n22 - 3n + 1Substituting, n = 1, 2, 3, 4, 5, ............., n in the above identity, we get                     133 - 033 = 3 . 122 - 3 ∙ 1 + 1                     233 - 133 = 3 . 222 - 3 ∙ 2 + 1

We will discuss here how to find the sum of the squares of first n natural numbers.Let us assume the required sum = STherefore, S = 122 + 222 + 322 + 422 + 522 + ................... + n22Now, we will use the below identity to find the value of S:n33 - (n - 1)33 = 3n22 - 3n + 1Substituting, n = 1, 2, 3, 4, 5, ............., n in the above identity, we get                     133 - 033 = 3 . 122 - 3 ∙ 1 + 1                     233 - 133 = 3 . 222 - 3 ∙ 2 + 1                     333 - 233 = 3 . 322 - 3 ∙ 3 + 1

We will discuss here how to find the sum of the squares of first n natural numbers.Let us assume the required sum = STherefore, S = 122 + 222 + 322 + 422 + 522 + ................... + n22Now, we will use the below identity to find the value of S:n33 - (n - 1)33 = 3n22 - 3n + 1Substituting, n = 1, 2, 3, 4, 5, ............., n in the above identity, we get                     133 - 033 = 3 . 122 - 3 ∙ 1 + 1                     233 - 133 = 3 . 222 - 3 ∙ 2 + 1                     333 - 233 = 3 . 322 - 3 ∙ 3 + 1                     433 - 333 = 3 . 422 - 3 ∙ 4 + 1

We will discuss here how to find the sum of the squares of first n natural numbers.Let us assume the required sum = STherefore, S = 122 + 222 + 322 + 422 + 522 + ................... + n22Now, we will use the below identity to find the value of S:n33 - (n - 1)33 = 3n22 - 3n + 1Substituting, n = 1, 2, 3, 4, 5, ............., n in the above identity, we get                     133 - 033 = 3 . 122 - 3 ∙ 1 + 1                     233 - 133 = 3 . 222 - 3 ∙ 2 + 1                     333 - 233 = 3 . 322 - 3 ∙ 3 + 1                     433 - 333 = 3 . 422 - 3 ∙ 4 + 1                     ......................................

We will discuss here how to find the sum of the squares of first n natural numbers.Let us assume the required sum = STherefore, S = 122 + 222 + 322 + 422 + 522 + ................... + n22Now, we will use the below identity to find the value of S:n33 - (n - 1)33 = 3n22 - 3n + 1Substituting, n = 1, 2, 3, 4, 5, ............., n in the above identity, we get                     133 - 033 = 3 . 122 - 3 ∙ 1 + 1                     233 - 133 = 3 . 222 - 3 ∙ 2 + 1                     333 - 233 = 3 . 322 - 3 ∙ 3 + 1                     433 - 333 = 3 . 422 - 3 ∙ 4 + 1                     ......................................              n33 - (n - 1)