I@Apu
INote for OP
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RAm gm laga k try kia lekin usse kuch ban hi nhi raha aisa ;-;
R$$2^{2n+1}>1+(2n+1)2^{n}$$
TReal potato
SMathematical Induction OP for these things
RAh i see.. i dont know anything about abt that chapter lol
SThe question said "n is an integer" that's why I suggested that.
SJust do differentiation or smth.
TSirLancelotDuLac
SAh nvm it says positive integers.
SShow function 2^(2x+1)-1-(2x+1).2^x is monotonous and then f(0)=0
SSo we're done.
RLol was it that easy.. i was thinking of it as some sort of series and finding general term cause i found this plm in my sns notebook
SYea you can do it that way too.
Ri cant see any series here ;-;
SYea, I'll do this by sns and send the answer (if possible, I don't have paper on me rn)
Rsure !! No hurry
Rthanks
JAgar sns se solve ho gaya toh कृप्या mujhe bhi ping kar dena
This looks interesting
This looks interesting
Ryo lance any updates on this?
SCouldn't do it by series, but one way I approached it initially was derivatives.
S$\frac{2^{2n+1}-1}{2n+1}>2^{n}$
TSirLancelotDuLac
SThis is the slope of secant from 0 to (2n+1).
_@Real potato @SirLancelotDuLac
_If we say F(x) = 2^n
From graph, as you mentioned the slope of secant, since the function 2^n is strictly increasing, the functional value will always be less than the average value of slope in the interval (0,2n+1).
What I mean to say is,
(F(2n+1) - F(0))/(2n+1-0) > F(x), where x belongs to (0,2n+1)
Therefore, 2^(2n+1) - 1/2n+1 > 2^n
From graph, as you mentioned the slope of secant, since the function 2^n is strictly increasing, the functional value will always be less than the average value of slope in the interval (0,2n+1).
What I mean to say is,
(F(2n+1) - F(0))/(2n+1-0) > F(x), where x belongs to (0,2n+1)
Therefore, 2^(2n+1) - 1/2n+1 > 2^n
_It's more of a calculus question?
Although there could be some method with SNS
Although there could be some method with SNS
