#### Thank you for registering.

One of our academic counsellors will contact you within 1 working day.

Click to Chat

1800-1023-196

+91-120-4616500

CART 0

• 0
MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

There are no items in this cart.
Continue Shopping

# 2+7+14+23+.... then find the sum upto 30 terms

AUREA
61 Points
5 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= ∑n+ 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.;)
2061 Points
5 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
2061 Points
5 years ago
i think @ AUREA is it incorrect because
AUREA forgot  one term in the  APPLIED AGP
pl correct
2061 Points
5 years ago
22-2 =
2
32-2 =
7
42-2
14
52 -2=
23 and so on

approve if useful

AUREA
61 Points
5 years ago
@akshat I think it should be S= ∑{(n+1)2-2}
which simplifies to ∑(n2+2n-1)
AUREA
61 Points
5 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 ; )
2061 Points
5 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