Prove, by method of induction, for all $n \in N$: $2 + 3.2 + 4.2^2 + \ldots + (n+1)2^{n-1} = n. 2^n$.

Prove, by method of induction, for all n ∈ N: 2 + 3.2 + 4.2² + … + (n+1)2ⁿ⁻¹ = n. 2ⁿ.

MISCELLANEOUS EXERCISE 4 (II) · Q96

n \in N

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✓ Free question

Let $P(n):2+3.2+4.2^2+\cdots+(n+1)2^{n-1}=n.2^n$ . Base: $n=1$ : L.H.S. $=2$ , R.H.S. $=1(2)=2$ ; holds. Hypothesis: assume true for $k$ : sum $=k.2^k$ . Step: add $(k+2)2^k$ : $k.2^k+(k+2)2^k=(2k+2)2^k=2(k+1)2^k=(k+1)2^{k+1}$ , matching the target $n.2^n$ at $n=k+1$ . Conclusion: true for all $n\in N$ .

✓Final answer

2+3.2+4.2^2+\cdots+(n+1)2^{n-1}=n.2^n

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Q.Prove, by method of induction, for all $n \in N$ : $2 + 3.2 + 4.2^2 + \ldots + (n+1)2^{n-1} = n. 2^n$ .

Concept understanding — Principle of Mathematical Induction

Q.Prove, by method of induction, for all n∈Nn \in Nn∈N: 2+3.2+4.22+…+(n+1)2n−1=n.2n2 + 3.2 + 4.2^2 + \ldots + (n+1)2^{n-1} = n. 2^n2+3.2+4.22+…+(n+1)2n−1=n.2n.

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Q.Prove, by method of induction, for all $n \in N$ : $2 + 3.2 + 4.2^2 + \ldots + (n+1)2^{n-1} = n. 2^n$ .

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