Binomial Theorem for Positive Integral Index

IMPORTANT

Binomial Theorem for Positive Integral Index: Overview

This topic covers concepts, such as, Finding Remainder Using Binomial Theorem, Finding Last Digit, Binomial Theorem & Problems Related to Binomial Expansion (Sqrt(a) + b)^n etc.

Important Questions on Binomial Theorem for Positive Integral Index

EASY
IMPORTANT

State whether the following binomial coefficient exist or not. If exists, find the value.

C515

EASY
IMPORTANT

State whether the following binomial coefficient exist or not. If exists, find the value.

C720

EASY
IMPORTANT

State whether the following binomial coefficient exist or not. If exists, find the value.

C910

EASY
IMPORTANT

State whether the following binomial coefficient exist or not. If exists, find the value.

C68

EASY
IMPORTANT

State whether the following binomial coefficient exist or not. If exists, find the value.

C812

MEDIUM
IMPORTANT

If the digits at ten's and hundred's places in 112016 are x and y respectively, then the ordered pair x,y is equal to

EASY
IMPORTANT

If the coefficient of 7th and 13th terms in the expansion of 1+xn, nN are equal, then n is equal to

MEDIUM
IMPORTANT

In the expansion of 1+x15+ 1+x16+ 1+x17+...+ 1+x30, the coefficient of x10 is

MEDIUM
IMPORTANT

The coefficient of x4 in the expansion of x2-3x210 is

HARD
IMPORTANT

The coefficient of xn in the expansion of 1+x2n and 1+x2n-1 are in the ratio

EASY
IMPORTANT

Given positive integers r>1, n>2 and the coefficient of 3rth & r+2th terms in the binomial expansion of 1+x2n are equal. Then,

EASY
IMPORTANT

In the expansion of x-1x6, the coefficient of x0 is

HARD
IMPORTANT

If n-1Cr=k2-3·nCr+1, then the value of k lies in

MEDIUM
IMPORTANT

If A and B are coefficient of xn in the expansions of 1+x2n and 1+x2n-1 respectively, then AB equals

MEDIUM
IMPORTANT

The value of 21C1-10C1+21C2-10C2+21C3-10C3+21C4-10C4++21C10-10C10 is

HARD
IMPORTANT

The number of terms in x 3 + 1 + 1 x 3 1 0 0 is -

EASY
IMPORTANT

The total number of dissimilar  terms in the expansion of x+a100+x-a100 after simplification is

EASY
IMPORTANT

x5+10x4 a+40x3 a2+80x2 a3+80 xa4+32 a5 is equal to

EASY
IMPORTANT

6th term in expansion of 2x2-13x210 is

EASY
IMPORTANT

If 27999 is divided by 7, then the remainder is