Prove that \sum_{i=0}^{n} 2^{i} = 2^{n + 1} - 1 Use mathematical induction for this proof and discuss/explain each step.a) i) b_0=1, b_1=2, b_n=2^nb 0 ​ =1,b 1 ​ =2,b n ​ =2 n Let a_1=b_0=1, a_2=b_1=2, ..., a_{n+1}=b_n=2^na 1 ​ =b 0 ​ =1,a 2 ​ =b 1 ​ =2,...,a n+1 ​ =b n ​ =2 n We have the geometric progression with the a=1a=1 and the common ratio r=2.r=2. \displaystyle\sum_{i=0}^{n}2^i=S_{n+1}=\displaystyle\sum_{j=1}^{n+1}a_j=\dfrac{a(1-r^{n+1})}{1-r} i=0 ∑ n ​ 2 i =S n+1 ​ = j=1 ∑ n+1 ​ a j ​ = 1−r a(1−r n+1 ) ​ =\dfrac{1(1-2^{n+1})}{1-2}=2^{n+1}-1= 1−2 1(1−2 n+1 ) ​ =2 n+1 −1 ii) Let P(n)P(n) be the proposition that the sum \displaystyle\sum_{i=0}^{n}2^i i=0 ∑ n ​ 2 i is 2^{n+1}-1.2 n+1 −1. Basis Step P(0)P(0) is true because 2^0=1=2^{0+1}-1.2 0 =1=2 0+1 −1. Inductive Step For the inductive hypothesis we assume that P(k)P(k) holds for an arbitrary positive integer k.k. That is, we assume that \displaystyle\sum_{i=0}^{k}2^i=1+2+4+...+2^k=2^{k+1}-1 i=0 ∑ k ​ 2 i =1+2+4+...+2 k =2 k+1 −1 Under this assumption, it must be shown that P(k + 1)P(k+1) is true, namely, that 1+2+4+...+2^k+2^{k+1}=2^{(k+1)+1}-11+2+4+...+2 k +2 k+1 =2 (k+1)+1 −1 is also true. When we add 2^{k+1}2 k+1 to both sides of the equation in P(k),P(k), we obtain 1+2+4+...+2^k+2^{k+1}=2^{k+1}-1+2^{k+1}1+2+4+...+2 k +2 k+1 =2 k+1 −1+2 k+1 =2\cdot2^{k+1}-1=2⋅2 k+1 −1 =2^{(k+1)+1}-1=2 (k+1)+1 −1 This last equation shows that P(k + 1)P(k+1) is true under the assumption that P(k)P(k) is true. This completes the inductive step. We have completed the basis step and the inductive step, so by mathematical induction we know that P(n)P(n) is true for all integers n\ge0n≥0 . That is, we have proven that \displaystyle\sum_{i=0}^{n}2^i=2^{n+1}-1 i=0 ∑ n ​ 2 i =2 n+1 −1a) i) b_0=1, b_1=2, b_n=2^nb 0 ​ =1,b 1 ​ =2,b n ​ =2 n Let a_1=b_0=1, a_2=b_1=2, ..., a_{n+1}=b_n=2^na 1 ​ =b 0 ​ =1,a 2 ​ =b 1 ​ =2,...,a n+1 ​ =b n ​ =2 n We have the geometric progression with the a=1a=1 and the common ratio r=2.r=2. \displaystyle\sum_{i=0}^{n}2^i=S_{n+1}=\displaystyle\sum_{j=1}^{n+1}a_j=\dfrac{a(1-r^{n+1})}{1-r} i=0 ∑ n ​ 2 i =S n+1 ​ = j=1 ∑ n+1 ​ a j ​ = 1−r a(1−r n+1 ) ​ =\dfrac{1(1-2^{n+1})}{1-2}=2^{n+1}-1= 1−2 1(1−2 n+1 ) ​ =2 n+1 −1 ii) Let P(n)P(n) be the proposition that the sum \displaystyle\sum_{i=0}^{n}2^i i=0 ∑ n ​ 2 i is 2^{n+1}-1.2 n+1 −1. Basis Step P(0)P(0) is true because 2^0=1=2^{0+1}-1.2 0 =1=2 0+1 −1. Inductive Step For the inductive hypothesis we assume that P(k)P(k) holds for an arbitrary positive integer k.k. That is, we assume that \displaystyle\sum_{i=0}^{k}2^i=1+2+4+...+2^k=2^{k+1}-1 i=0 ∑ k ​ 2 i =1+2+4+...+2 k =2 k+1 −1 Under this assumption, it must be shown that P(k + 1)P(k+1) is true, namely, that 1+2+4+...+2^k+2^{k+1}=2^{(k+1)+1}-11+2+4+...+2 k +2 k+1 =2 (k+1)+1 −1 is also true. When we add 2^{k+1}2 k+1 to both sides of the equation in P(k),P(k), we obtain 1+2+4+...+2^k+2^{k+1}=2^{k+1}-1+2^{k+1}1+2+4+...+2 k +2 k+1 =2 k+1 −1+2 k+1 =2\cdot2^{k+1}-1=2⋅2 k+1 −1 =2^{(k+1)+1}-1=2 (k+1)+1 −1 This last equation shows that P(k + 1)P(k+1) is true under the assumption that P(k)P(k) is true. This completes the inductive step. We have completed the basis step and the inductive step, so by mathematical induction we know that P(n)P(n) is true for all integers n\ge0n≥0 . That is, we have proven that \displaystyle\sum_{i=0}^{n}2^i=2^{n+1}-1 i=0 ∑ n ​ 2 i =2 n+1 −1