Question

Help please?

The value of the expression (3^(1001)+4^(1002))^2-(3^(1001)-4^(1002))^2 is k * 12^(1001) for some positive integer k. What is k?

Answer 1

`\bf \begin{array}{clclll} 3^(1001)&\qquad &4^(1002)\\ \uparrow &&\uparrow \\ a&&b \end{array}\\\\ -----------------------------\\\\ (a+b)^2-(a-b)^2\implies (a^2+2ab+b^2)-(a^2-2ab+b^2) \\\\\\ a^2+2ab+b^2-a^2+2ab-b^2\implies 2ab+2ab\implies 4ab \\\\\\ now\qquad 4(3^(1001))(4^(1002))=k12^(1001)\qquad \begin{cases} 12^(1001)\\ (4\cdot 3)^(1001)\\ 4^(1001)\cdot 3^(1001) \end{cases}`


`\bf \\\\\\ 4(3^(1001))(4^(1002))=k(4^(1001)\cdot 3^(1001))\implies \cfrac{4(3^(1001))(4^(1002))}{4^(1001)\cdot 3^(1001)}=k \\\\\\ 4\cdot \cfrac{3^(1001)}{3^(1001)}\cdot \cfrac{4^(1002)}{4^(1001)}=k\implies 4\cdot 4^(1002)4^(-1001)=k\impliedby \begin{array}{llll} \textit{same base}\\ \textit{add the}\\ exponents \end{array} \\\\\\ 4\cdot 4^(1002-1001)=k\implies 4\cdot 4^1=k\implies 16=k`