What is the value of 2^2 + 2*3^2 + 3*4^2 + 4*5^2 +... + 10*11^2 ?

What is the value of \(2^2+2∗3^2+3∗4^2+4∗5^2+...+10∗11^2\)?

A. 2850
B. 3600
C. 3850
D. 4530
E. 6530

\(2^2+2∗3^2+3∗4^2+4∗5^2+...+10∗11^2\) has 10 terms, if it is written as
\(1*2^2+2∗3^2+3∗4^2+4∗5^2+...+10∗11^2\) n = 1 to 10
Sum \((1*2^2+2∗3^2+3∗4^2+4∗5^2+...+10∗11^2)\) = \(Σ(n(n+1)^2)\)

\(Σ(n(n+1)^2)\) = \(Σ(n(n^2+1+2n))\) = \(Σ(n^3 + n + 2n^2)\) = \(Σn^3 + Σn + 2Σn^2\)
\(Σn^3 = \frac{n^2(n+1)^2}{4}, Σn = \frac{n(n+1)}{2}, Σn^2 = \frac{n(n+1)(2n+1)}{6}\)

\(Σ(n^3 + n + 2n^2)\) = \(\frac{n^2(n+1)^2}{4} + \frac{n(n+1)}{2} + 2*\frac{n(n+1)(2n+1)}{6}\)
= \(\frac{10^2*11^2}{4} + \frac{10*11}{2} + \frac{2*10*11*21}{6}\)
= 3025 + 55 + 770
= 3850

OR

\(2^2+2∗3^2+3∗4^2+4∗5^2+...+10∗11^2\) has 11 terms, if it is written as
\(0*1^2+1*2^2+2∗3^2+3∗4^2+4∗5^2+...+10∗11^2\) n = 1 to 11
Sum \((0*1^2+1*2^2+2∗3^2+3∗4^2+4∗5^2+...+10∗11^2)\) = \(Σ((n-1)n^2)\)

\(Σ((n-1)n^2)\) = \(Σn^3 - Σn^2\) = \(\frac{n^2(n+1)^2}{4} - \frac{n(n+1)(2n+1)}{6}\)
= \(\frac{11^2*12^2}{4} - \frac{11*12*23}{6}\)
= 4356 - 506
= 3850

Answer C.

Let's evaluate
2^2 = 4
2*3^2 = 18
...
5*6^2 = 180
6*7^2 = 294
7*8^2 = 448
8*9^2 = 648
9*10^2 = 900
10*11^2 = 1210

Clearly, the later elements of this series heavily contribute to the sum of the series.

Let's take the last six elements and approximate the lower-bound sum:
180 + ~295 + ~450 + 650 + 900 + ~1200 = 3685

A. 2850 --> < 3685. No!
B. 3600 --> < 3685. No!
C. 3850 --> YES! Closest
D. 4530 --> >> 3685. No!
E. 6530 --> >>> 3685. No!

FINAL ANSWER IS (C)

Posted from my mobile device

split the number into two category one with only 2 common and one with 4 common and simplify it answer is 3850

Tn = n(n+1) ^2

Sn = Summation Tn

Sn = n(n+1) (n+2) (3n+5) / 12

n =10

Putting n = 10

Sn = 3850

Posted from my mobile device

For the series we can see that it is of the form n(n+1)².

We know that sum of series = Summation of last term.

So summation of n³ + 2n² + n will give us the sum of the series.

= Summation( n³ + 2n² + n )
= (n(n+1)/2)² + 2n(n+1)(2n+1)/6 + n(n+1)/2

Putting n = 10 we get,

55² + 770 + 55 = 3850.


Thank you.

Posted from my mobile device

The series is in the form of

n(n+1)^2.....where n=from 1-10

After expansion and introducing sum for each variable

Sum(n^3+2n^2+n)

=[n(n+1)]^2/2 + 2[n(n+1)(2n+1)]/6 + n(n+1)/2

Substitute n =10

Sum= 3850

OA:C

Posted from my mobile device

What is the value of \(2^2 + 2*3^2 + 3*4^2 + 4*5^2 +... + 10*11^2\)?

A. 2850
B. 3600
C. 3850
D. 4530
E. 6530

We can do this by summation and some formulas, but first I'll try to answer it within realms of GMAT, as GMAT does not test you with these formulas/methods.

(I) Number properties.

You are struck and do not know where to start. Remember, it is SUM and we can check for divisibility by 2 and 3 at least.
All the choices are multiple of 10, so divisible by 2 and 5. Let us, therefore, concentrate only on 3.
\(2^2 + 2*3^2 + 3*4^2 + 4*5^2 +... + 10*11^2\)
Terms which have 3 or 6 or 9 are divisible by 3, so let us check on the remaining
\(2^2 + 4*5^2 +7*8^2+ 10*11^2=(3-1)^2+(3+1)(6-1)^2+(6+1)(9-1)^2+(9+1)(12-1)^2\)
All will leave a remainder of 1 when divided by 3, so remainder = 1+1*1+1*1+1*1=4=3+1 or final remainder is 1.
Check the choices for remainder as 1..
A. 2850....2+8+5+0=15...Remainder =0
B. 3600....3+6+0+0=9...Remainder =0
C. 3850....3+8+5+0=16...Remainder =1..Possible
D. 4530....4+5+3+0=12...Remainder =0
E. 6530....6+5+3+0=14...Remainder =2

So, C is the answer..

(II) Use of formula as also shown by many others

(III) Use of formula but a bit simplified..

\(2^2 + 2*3^2 + 3*4^2 + 4*5^2 +... + 10*11^2\). Add square of first 11 numbers to it.=\(\frac{n(n+1)(2n+1)}{6}=\frac{11*12*23}{6}=506\)
\(2^2 + 2*3^2 + 3*4^2 + 4*5^2 +... + 10*11^2\)+\(1^2+2^2 + 3^2 + 4^2 + 5^2 +... + 11^2\)=\(1^2+(1+1)2^2 + (1+2)*3^2 + (1+3)*4^2 + (1+4)*5^2 +... + (1+10)*11^2\)
=> \(1+2*2^2 + 3*3^2 + 4*4^2 + 5*5^2 +... + 11*11^2\)=\(1^3+2^3 + 3^3 + 4^3 + 5^3 +... + 11^3\)
Now Cube of first n positive numbers = \((\frac{n(n+1)}{2})^2=(\frac{11(11+1)}{2})^2=66^2=4356\)

As we have added 506 to our initial sum, the sum we are looking for = \(4356-506=3850\)

Tn = n(n+1) ^2

Sn = Summation Tn

Sn = n(n+1) (n+2) (3n+5) / 12

n =10

Putting n = 10

Sn = 3850

Posted from my mobile device

long way
2^2 + 2*3^2 + 3*4^2 + 4*5^2 +... + 10*11^2
1(2^2)=4; 2(3^2)=18..;
48, 100, 180, 294, 448, 648, 900, 1210;

sum~=3850

ans (C)