Question

$<marquee>hello$
Help me in the above attachment..

$\huge\underline\bold \red{question- 8}$


thank you ☺

Answer 1

$\bold{\huge{\underline{Correct Question}}}$

$\bold{1.2+2.2+3.2+.........+n.{2}^{n}=(n-1) {2}^{n+1} + 2}$

$\bold{\huge{\underline{Define 'n' value of Equation:}}}$

Substitute the value of 1 in place of n

$\bold{1.2+....=(1-1) \times {2}^{(1+1)} + 2} \\ \\ \bold{0 \times {2}^{2} + 2} \\ \\ \bold{0 \times 4 + 2 = 0} \\ \\ \fbox{0 + 2 = 2} \\ \\ \ \boxed{ \: \: 2 \: \: positive \: \: which \: \: means \: true}$

Substitute the value of k+1 in place of n!!

$\bold{P(k)=1.2 + {2.2}^{2}+ {3.2}^{3}+ ...+k{2}^{k}= (k-1){2}^{(k+1)}+2 }$

Substitute the value of k in place of n!!

$\bold{P(k+1) 1.2 +{2.2}^{2} + {3.2}^{3} + .......+ {k.2}^{k} + (k+1){2}^{k+1}= (k+1-1) {2}^{k+2} +2}$

Here, From Equation 1 we gets;

$\bold{1.2 + {2.2}{3} + {3.2}^{3} + ...k{2}{^k} = (k-1){2}^{(k+1)} +2}$

Now... Adding on both sides:-

$\bold{(k+1){2}^{(k+1)}}$

$\bold{1.2 + {2.2}^{2} + {3.2}^{3} + ........{k.2}^{k}+ {(k+1)2}^{(k+1)}}$

$\bold{{(k-1).2}^{(k+1)} + 2 + {(k+1)2}^{(k+1)}}$

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

$\bold{{2}^{(k+1)}2k}$

$\bold{{2.2}^{(k+1)}k}$

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

Hence,

$\bf{Conclusion:-}$

Here,

P(k+1) is true putting value and also p(k) is true

$\bold{Note:-}$

According to the principle of mathematical induction, statement . Considering on Follow true for all real number.

Answer 2

Answer:

10 thanxxx = 10 thanxx + follow ..............

15 thanxx = 20 thanxx + follow ..