2+7+14+23+.... then find the sum upto 30 terms
2+7+14+23+.... then find the sum upto 30 terms
Sri Ramya , 9 Years ago
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Magical Mathematics[Interesting Approach]> 2+7+14+23+.... then find the sum upto 30 ...
7 Answers
AUREA
Last Activity: 9 Years ago
2, 7, 14, 23, 34 …........ is a series whose common difference is an AP (i.e 5, 7, 9, 11,….).
S= 2 + 7 + 14 + 23 + ……………. An-1 + An
S= 2 + 7 + 14+ 23+………………+An-1 +An
Now, subtracting the equations
0= 2 + { 5 + 7 +9 + 11 +…….+ (n-1)} - An
An= 2+ { (n-1)(5+n-2)}
On simplifying, we get
An = n2+2n-1
Using the summation operator on the series we get
∑An= ∑n2 + 2*∑n – ∑1
where,
∑n2=n(n+1)(2n+1)/6
∑n=n(n+1)/2 and
∑1=n
putting n=30 we get the required sum.;)
grenade
Last Activity: 9 Years ago
it is
22-2 + 32-2 + 42-2 till 312-2
we will get is as
(∑n2) -1* – 2(30) -1* is taken because n2 starts from 12+22
where
∑n2=n(n+1)(2n+1)/6
on substituting the values it will be
30*31*61/6 = 9455
so
9455-1-60
=
9394
approve if useful
grenade
Last Activity: 9 Years ago
i think @ AUREA is it incorrect because
AUREA forgot one term in the APPLIED AGP
pl correct
grenade
Last Activity: 9 Years ago
22-2 =
2
32-2 =
7
42-2
14
52 -2=
23 and so on
approve if useful
AUREA
Last Activity: 9 Years ago
@akshat I think it should be S= ∑{(n+1)2-2}
which simplifies to ∑(n2+2n-1)
AUREA
Last Activity: 9 Years ago
putting the value n=30
∑n2=9455
2∑n=930
finally the sum comes to be S=9455+930-30 ,i.e, S=10355
approve if the solution is helpful ; )
grenade
Last Activity: 9 Years ago
thanks aurea for correcting me
inspite of n=31 i took it as n=30
the answer would have been 10355 if had placed n=31
sorry
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