#### Thank you for registering.

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

Please check your email for login details.

Click to Chat

1800-5470-145

+91 7353221155

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

# for 2^2+4^2+6^2+....(2n)^2 /1^2+3^2+5^2+...(2n-1)^2 to exceed 1.01,the maximum value of n is___________. ans:150.......PLS EXPLAIN THIS...

Grade:12

## 1 Answers

jitender lakhanpal
62 Points
8 years ago

hi sriraman

take  2^2 common from numerator we get (1^2+2^2+3^2+....(n)^2

use the standard formula

in numerator add and subtract
2^2+4^2+6^2+....(2n)^2

again with some algebraic steps u will get expression for which you can use formula

and then u will get the inequality in n

2^2+4^2+6^2+....(2n)^2 /1^2+3^2+5^2+...(2n-1)^2  > 1.01

## ASK QUESTION

Get your questions answered by the expert for free