Someone please walk me through how to solve this LOL THANKS
1 answer:
Answer:
ok it would be the answer on the bottom right which is
Step-by-step explanation:
So basically you just divide 34 by 7 to see how many times 7 goes into 34 which is 4 times. 7 times 4 gives you 28 and now you need to find the remainder, which is what's left inside the square root symbol. You put the x to the 4th power outside. So subtract 28 from 34 and that gives you 6 so you're left with x to the power of 6th inside the square root.
You might be interested in
Answer:
1. The given series is
4 ⋅ 6 + 5 ⋅ 7 + 6 ⋅ 8 + ... + 4n( 4n + 2) =
For n=1
L.H.S=4.6=24
R.H.S=[4×5×15]÷6
=300÷6
=50
So, for n=1,
L.H.S≠ R.H.S
Since the given expression is true for n=1 ,
So , the given series is untrue.
we should replace R.H.S by=4(n+1)(n+2)(4n-3)²
2.
12+42+72+.......+(3 n -2)2=
For n=1,
L.H.S=12
R.H.S=1×(6-3-1)/2
=2/2
=1
As L.H.S≠ R.H.S
We should Replace R.H.S by [(3 n-1)(3 n-2)]2
3.The given sequence is
2+4+6+....+2n=n(n+1)
L.H.S
P(1)=2
R.H.S
1×(1+1)
=1×2
=2
( b) L.H.S
P(n)=2+4+6+.....2 k
This is an A.P having n terms.
tex]S_{n}=
= n(n+ 1)
R.H.S=n(n+1)
So, P(k)=k(k+1)
(c) P(k+1)=2+4+6+.......+2(k+1)
This is an A.P having (k+1) terms.
=(k+1)(k+2)
So, P(k+1)= (k+1)(k+2)