Question

Please answer the following question -

Answer 1

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T(n)=n×n!

now...

T(1)+T(2)+T(3)+T(4)+T(5)+.......+T(10)

now...

T(1)=1×1!=(2-1)×1!=2×1!-1×1!=2!-1!

T(2)=2×2!=(3-1)×2!=3×2!-2!×1=3!-2!

T(3)=3×3!=(4-1)×3!=4×3!-3!×1=4!-3!

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T(10)=10×10!=(11-1)×10!=11×10!-10!×1=11!-10!

now.. summing These all we get...

T(1)+T(2)+T(3)+T(4)+T(5)+.......+T(10)

=(2!-1!)+(3!-2!)+(4!-3!)+(5!-4!)+........+(11!-10!)

=11!-1!

=(11!-1) And

........xd......

$:\: \underline{\underline{\bf{\large\mathfrak{~hope ~this ~help~you~~}}}}$

Answer 2

Answer:

$\huge\pink{elloh \: \: bhai \: jii}$

Step-by-step explanation:

Statement 1: It does not give any clue about n. As I can pick any value of n & make their sum 3124.

e.g. for n=4 let the terms be t1=0, t2=1 t3=9, t4=3114 The summation is 3124.

You can in fact choose any value of n and terms & still get their summation as 3124.

INSUFFICIENT

And NO you cannot use the formula n(n+1)/2. It is a big mistake.

If you remember the formula is for summation shortcut for n consecutive positive integers starting from 1. Whereas here you have been given TERMS and not consecutive integers. Do not be confused with term notation of t1,t2,t3,t4,t5...tn with 1,2,3,4,5,6....n.

Statement 2: This statement also does not help as Arithmetic mean can be written as:-

4=(t1+t2+...tn)/n OR 4n=(t1+t2+...tn)

to solve this equation I need to know the summation of all the terms which I do not have. I even do not know how many terms are there so (t1+t2+...tn) is impossible to determine discretely.

Therefore INSUFFICIENT.

Now combine both. As mentioned in Statement 2, in order to solve the equation you need summation of all the terms which you have from Statement 1 i.e. 3124

Put this value in the equation derived in Statement 2 i.e.

4n = (t1+t2+...tn) OR 4n=3124 => now do not solve further just select C and move on as you know that only 1 value of n can satisfy this equation.

Therefore C.