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
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...
Sum of a rational number and an irrational number will be an irrational number.
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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