Algebra

Method of Proof by Induction

Example

To prove that 1²+2²+3²+⋯+𝑛²=𝑛(𝑛+1)(2𝑛+1)6 Assume 12+22+32+⋯+𝑛2=𝑛(𝑛+1)(2𝑛+1)6 12+22+32+⋯+𝑛2+(𝑛+1)2=𝑛(𝑛+1)(2𝑛+1)6+(𝑛+1)2  =𝑛(𝑛+1)(2𝑛+1)+6(𝑛+1)26  =(𝑛+1){𝑛(2𝑛+1)+6(𝑛+1)}6  =(𝑛+1){2𝑛2+7𝑛+6}6  =(𝑛+1)(𝑛+2)(2𝑛+3)6  =(𝑛')(𝑛'+1)(2𝑛'+1)6 12+22+32+⋯+𝑛'2=(𝑛')(𝑛'+1)(2𝑛'+1)6 where 𝑛' is written for 𝑛+1; It is thus proved that if the formula be true for 𝑛 it is also true for 𝑛+1. But the formula is true when 𝑛=2 or 3, as may be shewn by actual trial; therefore it is true when 𝑛=4; therefore also when 𝑛=5, and so on; therefore universally true. 233

Example

The same theorem proved by the method of Indeterminate coefficients. Assume 1+22+32+⋯+𝑛2=𝐴+𝐵𝑛+𝐶𝑛2+𝐷𝑛3+⋯ ∴1+22+32+⋯+𝑛2+(𝑛+1)2=𝐴+𝐵(𝑛+1)+𝐶(𝑛+1)2+𝐷(𝑛+1)3+⋯ therefore, by subtraction, (𝑛+1)²=𝐵+𝐶(2𝑛+1)+𝐷(3𝑛²+3𝑛+1)+⋯ 𝑛²+2𝑛+1=𝐵+𝐶(2𝑛+1)+𝐷(3𝑛²+3𝑛+1) writing no terms in this equation which contain higher powers of 𝑛 than the highest which occurs on the left-hand side, for the coefficients of such terms may be shewn to be separately equal to zero. Now equate the coefficients of like powers of 𝑛; thus 3𝐷=1 ∴ 𝐷=13 2𝐶+3𝐷=2 ∴ 𝐶=12, and 𝐴=0 𝐵+𝐶+𝐷=1 ∴ 𝐵=16 therefore the sum of the series is equal to 𝑛6+𝑛²2+𝑛³3=𝑛(𝑛+1)(2𝑛+1)6 234

Sources and References

https://archive.org/details/synopsis-of-elementary-results-in-pure-and-applied-mathematics-pdfdrive