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Answer 1

Step-by-step explanation:

The given series is an Arithmetico-Geometric series i.e. it is a series obtained by multiplying terms of an AP and a GP.

Consider LHS of the given series and let it be 'S'.

$\sf S = 1 \cdot 2 + 2\cdot2^2 + 3\cdot2^3 +... + (n-1)\cdot2^{n-1} + n\cdot2^n$

Let it be equation 1

Multiply both sides with 2

$\sf 2S = 1 \cdot 2^2 + 2\cdot2^3 + 3\cdot2^4 +... + (n-1)\cdot2^{n} + n\cdot2^{n+1}$

Let it be equation 2

Now subtract equation 1 from equation 2

$\small{\sf-S = 1\cdot2 +\bigg[ (2\cdot2^2 - 1\cdot 2^2) + (3\cdot2^3 - 2\cdot2^3) +...+ (n\cdot2^n - \{n-1\}\cdot2^{n})\bigg]- n\cdot2^{n+1}}$

$\small{\sf-S = 1\cdot2 +\bigg[ 1\cdot2^2 +1\cdot2^3 +...+ 1\cdot2^n \bigg]- n\cdot2^{n+1}}$

$\small{\sf-S = \underbrace{\sf(2 + 2^2 + 2^3 + ...+ 2^n)}_{GP\, with\, a\,=\,r\,=\,2}-n\cdot2^{n+1}}$

Sum of n terms of a GP is given by:

• $\boxed{\tt S_n = \dfrac{a(r^n-1)}{r-1}}$

$\small{\sf-S =\dfrac{2(2^n-1)}{2-1}-n\cdot2^{n+1}}$

$\small{\sf-S ={2(2^n-1)-n\cdot2^{n+1}}$

$\small{\sf-S =2^{n+1}-2-n\cdot2^{n+1}}$

$\small{\sf-S =2^{n+1}(-n+1)-2}$

$\small{\sf-S =-(n-1)2^{n+1}-2}$

Multiply both sides with -ve 1

$\small{\sf S =(n-1)2^{n+1}+2}$

Hence the required result is proved that:

$\boxed{\tt 1 \cdot 2 + 2\cdot2^2 + 3\cdot2^3 +... + (n-1)\cdot2^{n-1} + n\cdot2^n = (n-1)2^{n+1} + 2}$

Answer 2

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Let P(n):

1.2+2.22 +3.23+...+n.2" (n-1)2+1+2

For n = 1

1.2 = 0.2° +2

2=2

→ P(n) is true for n = 1

Let P(n) is true for n= k, so 1.2+2.22 + 3.23+.........+k.2" = (k-1)2k+1 +2....(1)

We have to show that,

{1.2 +2.2² + 3.23+.....+k.2k) + (k + 1)2k+1 + k2k+2 + 2

Now,

= (1.2+2.22 +3.2³++k.2}+(k+1)2k+1

= [(k-1)2k+1 +2]+(k+1)2k+1 =2k+1+(k-1+k+1)+2

= k2k+2+2

→ p(n) is true for n = k +1

p(n) is true for all n epsilon N by PMI

${ \pink{ \sf \: hope \: it \: helps \: u}}$