Xn i=1 i2 i = (n − 1)2n+1 + 2

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Using induction to prove

We need prove a statement $\sum_{i=0} ^ {n} i \times 2^i = (n - 1) \times 2^{n+1} + 2$

Proof:

for n = 1

LHS =

$\sum_{i=0} ^ {1} i \times 2^i = 2$

RHS =

$(n - 1) \times 2^{n+1} + 2 = 0 + 2 = 2$

$LHS = RHS$

The given statement is true for n= 1

Let the given statement is true for n = k (integer)

$\sum_{i=0} ^ {k} i \times 2^i = (k - 1) \times 2^{k+1} + 2$

Now, Let n = k +1

$\sum_{i=0} ^ {k+1} i \times 2^i = \sum_{i=0} ^ {k} i \times 2^i + ( k+1 ) \times 2^{k+1}$

$= (k - 1) \times 2^{k+1} + 2 + (k+1 ) \times 2^ {k+1}$

$= k \times 2^ {k+1} - 2^ {k+1} + 2 + k \times 2^{k+1} + 2^{k+1}$

$= 2k \times 2^ {k+1} + 2 = k \times 2 ^ {k+1+1} + 2$

$\sum_{i=0} ^ {k+1} i \times 2^i = (k+1 -1) \times 2^ {k+1+1} + 2$

The given statement is true for n = k+1. By the principle of mathematical induction, it was proved that

$\sum_{i=0} ^ {n} i \times 2^i = (n - 1) \times 2^{n+1} + 2$ Using induction to prove

We need prove a statement $\sum_{i=0} ^ {n} i \times 2^i = (n - 1) \times 2^{n+1} + 2$

Proof:

for n = 1

LHS =

\sum_{i=0} ^ {1} i \times 2^i = 2

RHS =

(n - 1) \times 2^{n+1} + 2 = 0 + 2 = 2

LHS = RHS

The given statement is true for n= 1

Let the given statement is true for n = k (integer)

\sum_{i=0} ^ {k} i \times 2^i = (k - 1) \times 2^{k+1} + 2

Now, Let n = k +1

\sum_{i=0} ^ {k+1} i \times 2^i = \sum_{i=0} ^ {k} i \times 2^i + ( k+1 ) \times 2^{k+1}

= (k - 1) \times 2^{k+1} + 2 + (k+1 ) \times 2^ {k+1}

$= k \times 2^ {k+1} - 2^ {k+1} + 2 + k \times 2^{k+1} + 2^{k+1}$

= 2k \times 2^ {k+1} + 2 = k \times 2 ^ {k+1+1} + 2

\sum_{i=0} ^ {k+1} i \times 2^i = (k+1 -1) \times 2^ {k+1+1} + 2

The given statement is true for n = k+1. By the principle of mathematical induction, it was proved that

$\sum_{i=0} ^ {n} i \times 2^i = (n - 1) \times 2^{n+1} + 2$ Using induction to prove

We need prove a statement $\sum_{i=0} ^ {n} i \times 2^i = (n - 1) \times 2^{n+1} + 2$

Proof:

for n = 1

LHS =

\sum_{i=0} ^ {1} i \times 2^i = 2

RHS =

(n - 1) \times 2^{n+1} + 2 = 0 + 2 = 2

LHS = RHS

The given statement is true for n= 1

Let the given statement is true for n = k (integer)

\sum_{i=0} ^ {k} i \times 2^i = (k - 1) \times 2^{k+1} + 2

Now, Let n = k +1

\sum_{i=0} ^ {k+1} i \times 2^i = \sum_{i=0} ^ {k} i \times 2^i + ( k+1 ) \times 2^{k+1}

= (k - 1) \times 2^{k+1} + 2 + (k+1 ) \times 2^ {k+1}

$= k \times 2^ {k+1} - 2^ {k+1} + 2 + k \times 2^{k+1} + 2^{k+1}$

= 2k \times 2^ {k+1} + 2 = k \times 2 ^ {k+1+1} + 2

\sum_{i=0} ^ {k+1} i \times 2^i = (k+1 -1) \times 2^ {k+1+1} + 2

The given statement is true for n = k+1. By the principle of mathematical induction, it was proved that

$\sum_{i=0} ^ {n} i \times 2^i = (n - 1) \times 2^{n+1} + 2$ Using induction to prove

We need prove a statement $\sum_{i=0} ^ {n} i \times 2^i = (n - 1) \times 2^{n+1} + 2$

Proof:

for n = 1

LHS =

\sum_{i=0} ^ {1} i \times 2^i = 2

RHS =

(n - 1) \times 2^{n+1} + 2 = 0 + 2 = 2

LHS = RHS

The given statement is true for n= 1

Let the given statement is true for n = k (integer)

\sum_{i=0} ^ {k} i \times 2^i = (k - 1) \times 2^{k+1} + 2

Now, Let n = k +1

\sum_{i=0} ^ {k+1} i \times 2^i = \sum_{i=0} ^ {k} i \times 2^i + ( k+1 ) \times 2^{k+1}

= (k - 1) \times 2^{k+1} + 2 + (k+1 ) \times 2^ {k+1}

$= k \times 2^ {k+1} - 2^ {k+1} + 2 + k \times 2^{k+1} + 2^{k+1}$

= 2k \times 2^ {k+1} + 2 = k \times 2 ^ {k+1+1} + 2

\sum_{i=0} ^ {k+1} i \times 2^i = (k+1 -1) \times 2^ {k+1+1} + 2

The given statement is true for n = k+1. By the principle of mathematical induction, it was proved that

$\sum_{i=0} ^ {n} i \times 2^i = (n - 1) \times 2^{n+1} + 2$ Using induction to prove

We need prove a statement $\sum_{i=0} ^ {n} i \times 2^i = (n - 1) \times 2^{n+1} + 2$

Proof:

for n = 1

LHS =

\sum_{i=0} ^ {1} i \times 2^i = 2

RHS =

(n - 1) \times 2^{n+1} + 2 = 0 + 2 = 2

LHS = RHS

The given statement is true for n= 1

Let the given statement is true for n = k (integer)

\sum_{i=0} ^ {k} i \times 2^i = (k - 1) \times 2^{k+1} + 2

Now, Let n = k +1

\sum_{i=0} ^ {k+1} i \times 2^i = \sum_{i=0} ^ {k} i \times 2^i + ( k+1 ) \times 2^{k+1}

= (k - 1) \times 2^{k+1} + 2 + (k+1 ) \times 2^ {k+1}

$= k \times 2^ {k+1} - 2^ {k+1} + 2 + k \times 2^{k+1} + 2^{k+1}$

= 2k \times 2^ {k+1} + 2 = k \times 2 ^ {k+1+1} + 2

\sum_{i=0} ^ {k+1} i \times 2^i = (k+1 -1) \times 2^ {k+1+1} + 2

The given statement is true for n = k+1. By the principle of mathematical induction, it was proved that

$\sum_{i=0} ^ {n} i \times 2^i = (n - 1) \times 2^{n+1} + 2$ Using induction to prove

We need prove a statement $\sum_{i=0} ^ {n} i \times 2^i = (n - 1) \times 2^{n+1} + 2$

Proof:

for n = 1

LHS =

\sum_{i=0} ^ {1} i \times 2^i = 2

RHS =

(n - 1) \times 2^{n+1} + 2 = 0 + 2 = 2

LHS = RHS

The given statement is true for n= 1

Let the given statement is true for n = k (integer)

\sum_{i=0} ^ {k} i \times 2^i = (k - 1) \times 2^{k+1} + 2

Now, Let n = k +1

\sum_{i=0} ^ {k+1} i \times 2^i = \sum_{i=0} ^ {k} i \times 2^i + ( k+1 ) \times 2^{k+1}

= (k - 1) \times 2^{k+1} + 2 + (k+1 ) \times 2^ {k+1}

$= k \times 2^ {k+1} - 2^ {k+1} + 2 + k \times 2^{k+1} + 2^{k+1}$

= 2k \times 2^ {k+1} + 2 = k \times 2 ^ {k+1+1} + 2

\sum_{i=0} ^ {k+1} i \times 2^i = (k+1 -1) \times 2^ {k+1+1} + 2

The given statement is true for n = k+1. By the principle of mathematical induction, it was proved that

$\sum_{i=0} ^ {n} i \times 2^i = (n - 1) \times 2^{n+1} + 2$ Using induction to prove

We need prove a statement $\sum_{i=0} ^ {n} i \times 2^i = (n - 1) \times 2^{n+1} + 2$

Proof:

for n = 1

LHS =

\sum_{i=0} ^ {1} i \times 2^i = 2

RHS =

(n - 1) \times 2^{n+1} + 2 = 0 + 2 = 2

LHS = RHS

The given statement is true for n= 1

Let the given statement is true for n = k (integer)

\sum_{i=0} ^ {k} i \times 2^i = (k - 1) \times 2^{k+1} + 2

Now, Let n = k +1

\sum_{i=0} ^ {k+1} i \times 2^i = \sum_{i=0} ^ {k} i \times 2^i + ( k+1 ) \times 2^{k+1}

= (k - 1) \times 2^{k+1} + 2 + (k+1 ) \times 2^ {k+1}

$= k \times 2^ {k+1} - 2^ {k+1} + 2 + k \times 2^{k+1} + 2^{k+1}$

= 2k \times 2^ {k+1} + 2 = k \times 2 ^ {k+1+1} + 2

\sum_{i=0} ^ {k+1} i \times 2^i = (k+1 -1) \times 2^ {k+1+1} + 2

The given statement is true for n = k+1. By the principle of mathematical induction, it was proved that

$\sum_{i=0} ^ {n} i \times 2^i = (n - 1) \times 2^{n+1} + 2$ Using induction to prove

We need prove a statement $\sum_{i=0} ^ {n} i \times 2^i = (n - 1) \times 2^{n+1} + 2$

Proof:

for n = 1

LHS =

\sum_{i=0} ^ {1} i \times 2^i = 2

RHS =

(n - 1) \times 2^{n+1} + 2 = 0 + 2 = 2

LHS = RHS

The given statement is true for n= 1

Let the given statement is true for n = k (integer)

\sum_{i=0} ^ {k} i \times 2^i = (k - 1) \times 2^{k+1} + 2

Now, Let n = k +1

\sum_{i=0} ^ {k+1} i \times 2^i = \sum_{i=0} ^ {k} i \times 2^i + ( k+1 ) \times 2^{k+1}

= (k - 1) \times 2^{k+1} + 2 + (k+1 ) \times 2^ {k+1}

$= k \times 2^ {k+1} - 2^ {k+1} + 2 + k \times 2^{k+1} + 2^{k+1}$

= 2k \times 2^ {k+1} + 2 = k \times 2 ^ {k+1+1} + 2

\sum_{i=0} ^ {k+1} i \times 2^i = (k+1 -1) \times 2^ {k+1+1} + 2

The given statement is true for n = k+1. By the principle of mathematical induction, it was proved that

$\sum_{i=0} ^ {n} i \times 2^i = (n - 1) \times 2^{n+1} + 2$ Using induction to prove

We need prove a statement $\sum_{i=0} ^ {n} i \times 2^i = (n - 1) \times 2^{n+1} + 2$

Proof:

for n = 1

LHS =

\sum_{i=0} ^ {1} i \times 2^i = 2

RHS =

(n - 1) \times 2^{n+1} + 2 = 0 + 2 = 2

LHS = RHS

The given statement is true for n= 1

Let the given statement is true for n = k (integer)

\sum_{i=0} ^ {k} i \times 2^i = (k - 1) \times 2^{k+1} + 2

Now, Let n = k +1

\sum_{i=0} ^ {k+1} i \times 2^i = \sum_{i=0} ^ {k} i \times 2^i + ( k+1 ) \times 2^{k+1}

= (k - 1) \times 2^{k+1} + 2 + (k+1 ) \times 2^ {k+1}

$= k \times 2^ {k+1} - 2^ {k+1} + 2 + k \times 2^{k+1} + 2^{k+1}$

= 2k \times 2^ {k+1} + 2 = k \times 2 ^ {k+1+1} + 2

\sum_{i=0} ^ {k+1} i \times 2^i = (k+1 -1) \times 2^ {k+1+1} + 2

The given statement is true for n = k+1. By the principle of mathematical induction, it was proved that

$\sum_{i=0} ^ {n} i \times 2^i = (n - 1) \times 2^{n+1} + 2$

Xn i=1 i2 i = (n − 1)2n+1 + 2

Here's the Solution to this Question

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Question ID: mtid-5-stid-8-sqid-4054-qpid-2753