2 N 3 4N 1. Web solution for t6 =n(4n+1)−(n−1)(4n−3)=4n2+n−(4n2−3n−4n+3)=4n2+n−4n2+7n−3=8n−3=8(6)−3=48−3=45. Web see a solution process below:
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Web see a solution process below: Step 1 :equation at the end of step 1 : First, use this rule of exponents to eliminate the negative exponent on the leftmost term: Check that the middle term is two times the product of the numbers being squared in the first term and third. Step 3 :pulling out like terms : Web equation at the end of step 1 : We have 1 3 = 1 2. Xa = x−a1 (21)−1 + (21)0 +(21)1 ⇒. 12 + 22 + 32 + 42 +.+ n2 = (𝑛(𝑛 + 1)(2𝑛 + 1))/6 proving. Web example 1 for all n ≥ 1, prove that 12 + 22 + 32 + 42 +…+ n2 = (n(n+1)(2n+1))/6 let p(n) :
First, use this rule of exponents to eliminate the negative exponent on the leftmost term: Web equation at the end of step 1 : Web solution for t6 =n(4n+1)−(n−1)(4n−3)=4n2+n−(4n2−3n−4n+3)=4n2+n−4n2+7n−3=8n−3=8(6)−3=48−3=45. Web see a solution process below: Xa = x−a1 (21)−1 + (21)0 +(21)1 ⇒. 12 + 22 + 32 + 42 +.+ n2 = (𝑛(𝑛 + 1)(2𝑛 + 1))/6 proving. We have 1 3 = 1 2. Check that the middle term is two times the product of the numbers being squared in the first term and third. Web example 1 for all n ≥ 1, prove that 12 + 22 + 32 + 42 +…+ n2 = (n(n+1)(2n+1))/6 let p(n) : Step 1 :equation at the end of step 1 : Step 3 :pulling out like terms :
