A MATHS 問題~唔該大家

呢條題目好熟面口 係唔係se果份野黎架

a)真係要你自己做 好麻煩 幫唔到你><
b)
1(2)+2(2 2次)+3(2 3次)+‧‧‧+98(2 98次)=97(2 99次)+2
LHS=1(2)+2(2^2)+3(2^3)+‧‧‧+98(2^98)
分開做 2+2(2)x2+3(2^2)x2+‧‧‧+98(2^97)x2
之後抽個2出黎
=2+2[2(2)+3(2^2)+‧‧‧+98(2^97)]
=2+2x97(2^98)
=2+2x97(2^99)
=RHS
so 1(2)+2(2 2次)+3(2 3次)+‧‧‧+98(2 98次)=97(2 99次)+2


希望幫到你^^

我唔記(a)啦，就咁用答案算
2(2)+3(2^2)+4(2^3)+......(n+1)(2^n)=n[2^(n+1)]

(b)1(2)+2(2^2)+3(2^3)+......98(2^98)
=(2-1)(2)+(3-1)(2^2)+(4-1)(2^3)+......+(99-1)(2^98)
=2(2)+3(2^2)+4(2^3)+......+99(2^98)-[1(2)+1(2^2)+1(2^3)+......+(1(2^98)]
=98(2^99)-{[1(2)(1-2^98)]/(1-2)} (By A.P. Sum)
=98(2^99)-2(2^98-1)
=98^(2^99)-2^99+2
=(98-1)(2^99)+2
=97(2^99+2)

P.S. A.P. Sum=[a(1-r^n)]/(1-r)

a)prove ,by M.I.,that
2(2)+3(2^2)+4(2^3)+‧‧‧+(n+1)(2^n) = n 2^(n+1) for all positive integers n
(1) 2^(1+1) = 4 = 2(2)
suppose 2(2)+3(2^2)+4(2^3)+‧‧‧+(k+1)(2^k) = k 2^(k+1)
consider
2(2)+3(2^2)+4(2^3)+‧‧‧+ (k+1)(2^k) + (k+2) 2^(k+1)
= k 2^(k+1) + (k+2) 2^(k+1)
= (k + k + 2) 2^(k+1)
= 2 (k + 1) 2^(k+1)
= (k+1) 2^(k+1+1)
so, when case n = k is true, case n = k+1 is also true
--------------------------------------------------------------------------------------

b)
1(2)+2(2^2)+3(2^3)+‧‧‧+98(2^98)
= 2(2)+3(2^2)+4(2^3)+‧‧‧+(99)(2^98) - [(2)+(2^2)+(2^3)+‧‧‧+(2^98)]
= (98) 2^99 - (2)(2^98 - 1)/(2-1)
= (98) 2^99 - 2^99 + 2
= (97) (2^99) + 2

容易

不明白再 e-mail 問我啦 ^_^


　2(2)+3(2 2次)+4(2 3次) + ‧‧‧ + (n+1)(2 n次)

= (1+1) (2) + (1+2) (2 2次) + (1+3) (2 3次) + ‧‧‧ + (1+n) (2 n次)


很大的提示吧