Derivation of 1×2+2×3+3×4+....…+n(n+1) = [n(n+1)(n+2)]/3

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Sum of cubes of the first n natural numbers - Formula derivation

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Hello, Everyone. In this, we are going to derive the formula to find the sum of cubes of the first n natural numbers. Let S = 1 3  + 2 3  + 3 3  + .......... + (n-1) 3  + n 3 (m+1) 4  - m 4  = (m 4  + 4m 3  + 6m 2 + 4m + 1) - m 4  = 4m 3  + 6m 2 + 4m + 1 ------------ (i) put m = 1,2,3, ....... , (n-1), n in (i) Then, 2 4  - 1 4   = 4(1) 3  + 6(1) 2  + 4(1) + 1 ------------ (1) 3 4  - 2 4   = 4(2) 3  + 6(2) 2  + 4(2) + 1 ------------ (2) 4 4  - 3 4   = 4(3) 3  + 6(3) 2  + 4(3) + 1 ------------ (3) ................... n 4  - (n-1) 4   = 4(n-1) 3  + 6(n-1) 2  + 4(n-1) + 1 ------------ (n-1) (n+1) 4  - n 4   = 4(n) 3  + 6(n) 2  + 4(n) + 1 ------------ (n) Now, by adding equations (1),(2),(3), ...... ,(n-1) and (n), we get the following equation. (n+1) 4...
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Sum of a rational number and an irrational number will be an irrational number.

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Hello Everyone. Before going to the proof, let us go into the basics of Rational and Irrational numbers. Rational Number  :   A number that can be expressed as the fraction p/q of two integers p and q, where q not equal to zero. Irrational Number  :   A number that cannot be expressed as the fraction p/q of two integers p and q,  where q not equal to zero. What is required to prove  :   Sum of a rational number and an irrational number will be an irrational number. Proof  : Let "m" be an irrational number. --------- (1) Let "n" be a rational number. ------------(2) Let us think that we don't know whether m + n is rational or irrational number. So, let us assume m + n as a rational number initially. Then, m + n = p/q where p and q are some integers and q is not equal to zero. (According to the definition) So, m + n = p/q Subtract "n" on both sides. Then, m = (p/q) - n = difference of two rational numbers Difference of two rational numbers will be a...
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