What is the value of (2^4−1)/(2−1) + (3^4−1)/(3−1) + (4^4−1)/(4−1) +

Question

Competition Mode Question

What is the value of  \frac{(2^4−1)}{(2−1)} + \frac{(3^4−1)}{(3−1)} + \frac{(4^4−1)}{(4−1)} + ... + \frac{(10^4−1)}{(10−1)} = ?

A. 3,462 
B. 3,471 
C. 3,479 
D. 3,581 
E. 4,022

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nick1816 · Answer

n^4-1 = (n^2-1)(n^2+1) = (n+1)(n-1)(n^2+1)

\frac{n^4-1}{n-1} = (n^2+1)(n+1)

\frac{n^4-1}{n-1} = (n^3+n^2+n+1)

Σ\frac{n^4-1}{n-1} = Σn^3+Σn^2+Σn+Σ1

Σ\frac{n^4-1}{n-1} =  ^2+\frac{n(n+1)(2n+1)}{6}+\frac{n(n+1)}{2}+n

Put n=10

=  - 4

= 55(55+7+1) +6

= 55*63+6

=63(50)+63*5+6

=3150+315+6

=3471

freedom128 · Answer

(a^4−1)/(a−1)= (a+1)(a^2+1)

a=2.. (2+1)(2^2+1) = 3.5 = 15
a=3.. (3+1)(3^2+1) = 4.10 = 40
a=4.. (4+1)(4^2+1) = 5.17 = 85
a=5.. (5+1)(5^2+1) = 6.26 = 156
a=6.. (6+1)(6^2+1) = 7.37 = 259
a=7.. (7+1)(7^2+1) = 8.50 = 400
a=8.. (8+1)(8^2+1) = 9.65 = 585
a=9.. (9+1)(9^2+1) =10.82 = 820
a=10.. (10+1)(10^2+1) = 1111

Sum from a=2 to a=10 is
= 15 + 40 + 85 + ... + 1111
= 3471

FINAL ANSWER IS (B)

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yashikaaggarwal · Answer

(2^4-1)/(2-1) can be written as (2^2+1^2)(2^2-1^2)/(2-1)
(A^2-B^2) property
(2^2+1^2)(2^2-1^2)/(2-1) can be further bifurcated into (2^2+1^2)(2+1)(2-1)/(2-1)

Therefore the equation can be written as.
(2^2+1^2)(2+1)+(3^2+1^2)(3+1)+.........+(10^2+1^2)(10+1)
5(3)+10(4)+.........+(101)(11)
15+40+85+......+1111
3471

OA is B

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suchita2409 · Answer

This is my way of solving.The method can be wrong.
2^4 - 1 /(2-1) can also be written as(( 2^2+1)(2+1)(2-1))/(2-1)

"using formula 
a^2 - b^2 =(a+b)(a-b)"

= (2^2+1)(2+1) 
 
the pattern gives units digit to be 0 + 0 + 5 + 6 + 3 + 0+ 5+ 0 = 9

the only option that gives 9 at units place is C.

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GMATWhizTeam · Answer

Solution 
 • The nth term(Let’s say  a_n ) of the given summation (Let’s say  S_n )  = \frac{(n^4 – 1)}{n-1}  , where n is an integer, and  2≤ n ≤10 
   ⟹ a_n = \frac{(n^2 +1)(n^2 -1)}{n-1} 
  ⟹ a_n = \frac{(n^2 +1)(n -1)(n+1)}{n-1} 
  ⟹ a_n = (n^2 +1)(n+1) = n^3 + n^2 +n +1  
• From the expression of nth term, we can observe that  S_n  = summation of the cube of each integer from 2 to 10 + summation of the square of each integer from 2 to 10 + summation of each integer from 2 to 10 + summation of 1 nine times
• Thus, we can write,  S_n = (\frac{n(n+1)}{2})^2 - 1^3 + \frac{n(n+1)(2n+1)}{6} – 1^2 + \frac{n(n+1)}{2}) – 1 + 9    2≤ n ≤10 , we have subtracted 1 to exclude it]
   ⟹ S_n = (\frac{10(11)}{2})^2 - 1^3 + \frac{10(11)(21)}{6} – 1^2 + \frac{10(11)}{2}) – 1 + 9 
  ⟹ S_n = 55^2 + 55*7 + 55 + 6 = 3471   
Thus, the correct answer is  Option B.

exc4libur · Answer

(2^4−1)/(2−1)
(2^2+1)(2^2-1)/(2-1)
(2^2+1)(2-1)(2+1)/(2-1)
(2^2+1)(2+1)
...
(10^2+1)(10+1)

sum: 3471

ans (B)

Saasingh · Answer

tn = (n⁴ - 1)/(n-1)
 = ((n²)²-1)/(n-1)
 = (n²-1)(n²+1)/(n-1)
 = (n+1)(n²+1)
 = n³ + n² + n + 1

So , sum of tn = sum of series.

Note that first term is created with n = 2. Therefore summation is from 2 to 10.

Sn = (n*(n+1)/2)² + (n*(n+1)(2n+1)/6) + (n*(n+1)/2) + 1 - 4 (-4 because summation is from 2 to 10 so subtract sumnation(1).

Putting n = 10, we get option A

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unraveled · Answer

What is the value of  \frac{(2^4−1)}{(2−1)}+\frac{(3^4−1)}{(3−1)}+\frac{(4^4−1)}{(4−1)}+...+\frac{(10^4−1)}{(10−1)} = ?

A. 3,462
B. 3,471
C. 3,479
D. 3,581
E. 4,022
 \frac{(2^4−1)}{(2−1)}+\frac{(3^4−1)}{(3−1)}+\frac{(4^4−1)}{(4−1)}+...+\frac{(10^4−1)}{(10−1)} 
 \frac{(2−1)(2+1)(2^2+1)}{(2−1)}+\frac{(3−1)(3+1)(3^2+1)}{(3−1)}+\frac{(4−1)(4+1)(4^2+1)}{(4−1)}+...+\frac{(10−1)(10+1)(10^2+1)}{(10−1)} 
 (2+1)(2^2+1)+(3+1)(3^2+1)+(4+1)(4^2+1)+...+(10+1)(10^2+1) 
3*5+4*10+5*17+6*26+.... +11*101 = 3471

Answer B.

*PS: The series can be written in the form  (n+2)  = n^3+4n^2+6n+4  where number of terms is 9(n=1 to n = 9) 
Then from  Σ(n^3+4n^2+6n+4)  =  Σn^3 + 4Σn^2 + 6Σn + 4  which sums to 3439.

Archit3110 · Answer

if we observe we have total 9 terms and for every even term eg 2^4-1/(2-1) ; the value would be odd and for every odd term eg 3^3-1/(3-1) value would be even
so answer option would be 5terms of odd + 4 terms of even i.e odd answer
option A & E are out

for given sequence the terms would be = 15+40+85+156+259+400+585+820+1111 ; 3471
OPTION B

What is the value of (2^4−1)/(2−1)+(3^4−1)/(3−1)+(4^4−1)/(4−1)+...+(10^4−1)/(10−1)=?

A. 3,462
B. 3,471
C. 3,479
D. 3,581
E. 4,022

nikitamaheshwari · Answer

why are we subtracting the summation of 1 here? If 1 is placed in the same format wouldnt the result be 0?
1^4-1/1-1?? Can you tell me the why are we subtracting -4 when the role of 1 here can be ignored

Bunuel  - Thanks

unraveled · Answer

nikitamaheshwari

n^3 + n^2 + n + 1  
n = 1 then 
 n^3 + n^2 + n + 1 = 1^3 + 1^2 + 1 + 1 = 4

Hoe this helps.

Kinshook · Answer

What is the value of  \frac{(2^4−1)}{(2−1)} + \frac{(3^4−1)}{(3−1)} + \frac{(4^4−1)}{(4−1)} + ... + \frac{(10^4−1)}{(10−1)} = ?

n^4 - 1 = (n^2 - 1)(n^2 + 1) = (n-1)(n+1)(n^2+1) =  
 x_n = \frac{(n^4 - 1)}{(n-1)} = (n+1)(n^2+1) = n^3 + n^2 + n + 1

\sum_{i=1}^{n} x_i = \sum_{i=1}^{n} (n^3 + n^2 + n + 1) = (\frac{n(n+1)}{2})^2 + \frac{n(n+1)(2n+1)}{6} + \frac{n(n+1)}{2} + n

\sum_{i=1}^{10} x_i = \sum_{i=1}^{10} (n^3 + n^2 + n + 1) - 4= (\frac{10(10+1)}{2})^2 + \frac{10(10+1)(2*10+1)}{6} + \frac{10(10+1)}{2} + 10 - 4 = 3025 + 385 + 55 + 10 - 4 = 3475 - 4 = 3471

4 is deducted for i=1;

IMO B

What is the value of (2^4−1)/(2−1) + (3^4−1)/(3−1) + (4^4−1)/(4−1) +

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What is the value of (2^4−1)/(2−1) + (3^4−1)/(3−1) + (4^4−1)/(4−1) +

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